定义

错误汇总

前缀和数组的原数组, 要考虑其全部的限制因素

有一个bool A[]的原数组 (取值0/1), 他对应一个int sum[]的前缀和数组;

我们有关系: A[l, l+1, ..., r] >= k, 即对应不等式: sum[r] - sum[l-1] >= k, 对应差分约束的有向边为: sum[l-1] + k <= sum[r] 即(l-1) -> (r) [w];

这一类关系: 即(l-1) -> (r) [w];

第二类关系 也容易考虑到, 即A[] >= 0, 对应(i-1) -> (i) [0];

但一定不要忘记, 还有一个限制, 即A[] <= 1, 故必须还要有: (i) -> (i-1) [-1], 否则就错了;

算法模板

//{ InequationSystem-Declaration
template< class _Type_Edge, class _Type_Distance>
class InequationSystem{
public:const _Type_Distance * Distance;//-- variable endsInequationSystem( const Graph_Weighted< _Type_Edge> *);bool Initialize();
private:const Graph_Weighted< _Type_Edge> * graph;int points_range;_Type_Distance * distance;bool * isInQueue;int * queue;int queue_head, queue_tail;int * pathLength;
};
//} InequationSystem-Declaration//{ InequationSystem-Implementation
//<
// @Brief
// . A edge `a->b (w)` in `graph`, means a inequation `Distance[a] + w <= Distance[b]` (`w` can be any-integer);
// . This algorithm use **SPFA** to get the **Maximal-Distance** for each point, finally `Distance[x] >= 0` if this Inequation-System is solvable
//   . More precisely, `Distance[x]` would be the **Minimal-Value (>=0)** under the whole Inequation-System;
// @Function( Initialize)
// . If return `true` when this Inequation-System is solvable (then the answer is `Distance[]`), otherwise return `false;
// @Hint
// . Cuz the core-algorithm is SPFA, if TimeOut occurs, you can use the next-code in `@Mark_0`;
//   . if( ++ timeOut > points_range * @Unfinished(5/10/15/...)){ return true;}
template< class _Type_Edge, class _Type_Distance> InequationSystem< _Type_Edge, _Type_Distance>::InequationSystem( const Graph_Weighted< _Type_Edge> * _graph):graph( _graph){distance = new _Type_Distance[ graph->Get_pointsCountMaximum()];Distance = distance;isInQueue = new bool[ graph->Get_pointsCountMaximum()];queue = new int[ graph->Get_pointsCountMaximum() + 1];pathLength = new int[ graph->Get_pointsCountMaximum()];
}
template< class _Type_Edge, class _Type_Distance> bool InequationSystem< _Type_Edge, _Type_Distance>::Initialize(){points_range = graph->Get_pointsCount();memset( distance, 0, sizeof( _Type_Distance) * points_range);memset( pathLength, 0, sizeof( int) * points_range);for( int i = 0; i < points_range; ++i){ queue[ i] = i;}memset( isInQueue, true, sizeof( bool) * points_range);queue_head = 0, queue_tail = points_range;//--int cur, nex, edge;while( queue_head != queue_tail){//>< @Mark_0cur = queue[ queue_head ++];isInQueue[ cur] = false;if( queue_head > points_range){ queue_head = 0;}for( edge = graph->Head[ cur]; ~edge; edge = graph->Next[ edge]){nex = graph->Vertex[ edge];if( distance[ cur] + graph->Weight[ edge] > distance[ nex]){pathLength[ nex] = pathLength[ cur] + 1;if( pathLength[ nex] >= points_range){return false;}distance[ nex] = distance[ cur] + graph->Weight[ edge];if( false == isInQueue[ nex]){isInQueue[ nex] = true;queue[ queue_tail ++] = nex;if( queue_tail > points_range){ queue_tail = 0;}}}}}return true;
}
//} InequationSystem-Implementation

例题

CSDN-129476594;

@Delimiter

经验总结

题型/算法分析

Cuz this kind of problem must corresponds to a Graph G G G (an edge a → b ( w ) a\to b \ (w) a→b (w) always denotes V [ a ] + w ≤ V [ b ] V[a] + w \leq V[b] V[a]+w≤V[b], then calculating the Minimal-Value for every V [ x ] V[x] V[x] under the premise V [ a l l ] ≥ 0 V[ all] \geq 0 V[all]≥0);
. Although the Normal-SPFA can solve the problem, but it maybe Out-Of-Time, that is, if there is better algorithms, we should always give up the idea of SPFA;
. According to the different characteristics of the Graph, there are the best-method to solving the problem :
+ If G G G is a D A G DAG DAG (whatever the edge-weight is, positive/zero/negative), see the below-section;
+ If all edges-weight of G G G are ≥ 1 \geq 1 ≥1, see the below-section;
+ If all edges-weight of G G G are ≥ 0 \geq0 ≥0, see the below-section;

若 G G G 为 D A G DAG DAG, 最优算法为拓扑序

Link

--

The optimal-method is:

ASSERT( the problem must be Consistent/Solvable);
V[] = 0;
for( c : $(the topological-sequence of G){ASSERT( V[c] is the answer of `c`);for( n : $(the adjacent-points of `c`)){w = the weight of the edge;V[ n] = max( V[n], max( V[c] + w, 0)); //< find the Maximal-Distance of `n` from any start-point in the DAG, and also ensure `V[all]` always `>= 0`;}
}

若所有边权 ≥ 1 \geq 1 ≥1, 最优算法为拓扑序

AcWing-1192. 奖金

G G G such that all edges-weight are ≥ 1 \geq 1 ≥1, an edge denotes an equality a + w ≤ b a + w \leq b a+w≤b where w ≥ 1 w \geq 1 w≥1;
+ The problem is Inconsistent if and only if G G G is not a DAG;
. Otherwise (i.e., Consistent), let D i s t [ x ] Dist[ x] Dist[x] be the Maximal-Distance from any start-point in the DAG to x x x; finally, the answer is D i s t Dist Dist (that is, the Minimal-Value under the condition D i s t [ x ] ≥ 0 Dist[x] \geq 0 Dist[x]≥0);

若所有边权 ≥ 0 \geq 0 ≥0, 最优算法为SCC

AcWing-368. 银河

G G G such that all edges-weight are ≥ 0 \geq 0 ≥0, an edge denotes an equality a + w ≤ b a + w \leq b a+w≤b where w ≥ 0 w \geq 0 w≥0;
+ The problem is Inconsistent if and only if There is an Non-Zero edge within a S C C SCC SCC of G G G;
. Otherwise (i.e., Consistent), let G 1 G1 G1 be the DAG after shrinking all S C C SCC SCCs of G G G, D i s t [ x ] Dist[ x] Dist[x] where x ∈ G 1 x \in G1 x∈G1 be the Maximal-Distance from any start-point in the DAG G 1 G1 G1 to x x x; finally, the answer of a a a equals to D i s t [ A ] Dist[ A] Dist[A] where A ∈ G 1 A \in G1 A∈G1 is the SCC of a ∈ G a \in G a∈G (that is, the Minimal-Value under the condition ∀ a ∈ G , ≥ 0 \forall a \in G, \geq 0 ∀a∈G,≥0);
. Note that, any edge a → b ( w ) a\to b (w) a→b(w) in G G G where a , b a,b a,b not belongs to the same S C C SCC SCC, should corresponds to an edges A → B ( w ) A\to B (w) A→B(w) in G 1 G1 G1 where A , B A,B A,B denotes the S C C SCC SCC; equivalently, we can choose a Maximal-Edge a → b ( w ) a\to b(w) a→b(w) as the unique-edge A → B A \to B A→B

@Delimiter( below are the old-interpretation)

If w ≥ 0 w \geq 0 w≥0 for all x i + w ≤ x j xi + w \leq xj xi+w≤xj, as we know, it forms a Directed-Graph, and we should calculate Greatest-Path with the origin D i s t [ a l l ] = 0 Dist[ all] = 0 Dist[all]=0;
1 Inconsistent (there exists a Positive-Circuit) ⇔ \Leftrightarrow ⇔ there exists a edge > 0 > 0 >0 in a SCC;
. Proof p → q p \to q p→q, a Positive-Circuit must belongs to a SCC;
. Proof q → p q \to p q→p, due to all edges ≥ 0 \geq 0 ≥0, a circuit that contains a > 0 >0 >0 edge in a SCC, the sum of this circuit must > 0 > 0 >0;
. As a lemma, if it is consistent, all edges in a SCC are 0 0 0.
2 The maximal-distance of a point a a a ⇔ \Leftrightarrow ⇔ the maximal-distance of a SCC (to which the point a a a belongs) in the DAG (after shrinking-SCC);
. Now, all edges in a SCC are 0 0 0 (so, the D i s t [ ] Dist[] Dist[] of all points in a SCC are the same), all edge in DAG ≥ 0 \geq 0 ≥0; the maximal-distance of a point a a a, must with the form s → . . . → a s \to ... \to a s→...→a, it is equivalent to its corresponding DAG-Path S → . . . → A S \to ... \to A S→...→A (cuz any edge within a SCC is 0 0 0); therefore, we calculate the Maximal-distance on a DAG;

Total cost: O ( n + m ) ∗ 2 O(n+m) * 2 O(n+m)∗2

Tarjan_SCC< int> tarjan( &G); //< all edges in `G` are `> 0`
tarjan.Initialize();
//{
for( int i = 0; i < n; ++i){for( int j, z = G.Head[ i]; ~z; z = G.Next[ z]){j = G.Vertex[ z];if( (tarjan.Scc_id[ i] == tarjan.Scc_id[ j]) && (G.Weight[ z] != 0)){cout<< "Inconsistent";return;}}
}
//}
auto DAG = tarjan.Build_DAG();
DAG_Tools< int> dag_tool( DAG);
dag_tool.IfIsDAG_andThen_getTopologySequence();
vector< int> Dist( DAG->Get_pointsCount(), 0);
for( int i = 0; i < DAG->Get_pointsCount(); ++i){int c = dag_tool.Topological_Sequence[ i];FOR_EDGES_( c, nex, wth, DAG->)Dist[ nex] = max( Dist[ c] + wth, Dist[ nex]);}
}for( int i = 0; i < n; ++i){Dist[ tarjan.Scc_id[ i]];  //< the answer of point `i`; (the minimal-value under the condition `all xi >=0 `
}

当答案要求为(满足 x i ≤ d xi \leq d xi≤d的最大值),而非通常的(满足 x i ≥ d xi \geq d xi≥d的最小值)

The usual question is calculating the minimal-value for every x i xi xi under the condition x i ≥ d ∀ x i xi \geq d \quad \forall xi xi≥d∀xi

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But when the question is that calculating the maximal-value for every x i xi xi under the condition x i ≤ d ∀ x i xi \leq d \quad \forall xi xi≤d∀xi, it also can be solved:

• Continue in the usual setting, calculating the minimal-value for every x i xi xi under the condition x i ≥ D ∀ x i xi \geq D \quad \forall xi xi≥D∀xi where D = − 1 ∗ d D = -1 * d D=−1∗d;
  supposing that the final answer denoted by X i Xi Xi;
• Return back to the current setting, the maximal-value for x i xi xi under the condition x i ≤ d ∀ x i xi \leq d \quad \forall xi xi≤d∀xi equals − 1 ∗ X i -1 * Xi −1∗Xi;

不等式恒为 ≤ \leq ≤

Whatever the problem is, making sure that every inequality always be the form of x i + w ≤ x j xi + w \leq xj xi+w≤xj which corresponds to a edge x i → x j xi \to xj xi→xj with weight w w w. (Any Inequalities-System problem can always be solved in this form)

负边不可省略

The Greatest-Path of a point may contains Negative-Edge (e.g., the edges on the path are 10, -1)

差分约束

问题描述

Given a sequence of variables x 1 , x 2 , . . . , x n x1, x2, ..., xn x1,x2,...,xn, and a Inequality-Group which consists of a set of relations with the form x i + c ≤ x j xi + c \leq xj xi+c≤xj where c c c is a constant.

The question is, find out the minimal value of every x i xi xi under the condition ( a l l x i ) ≥ d (all \ xi) \geq d (all xi)≥d (or find out the minimal sum of x i xi xi under the same condition, actually they are the same), (or find out the maximal value of every x i xi xi under the condition ( a l l x i ) ≤ d (all \ xi) \leq d (all xi)≤d, in fact they are also the same). A another view-point see Geometrical description

问题分析

It can be proved that the solution-set is either infinite or empty (no solution / inconsistent) .

For example
. x 1 + 1 ≤ x 2 x1 + 1 \leq x2 x1+1≤x2 implies the solutions . . . ( − 1 , ≥ 0 ) , ( 0 , ≥ 1 ) , ( 1 , ≥ 2 ) . . . ... (-1, \geq 0), (0, \geq 1), (1, \geq 2) ... ...(−1,≥0),(0,≥1),(1,≥2)... (abbr., ( x , > x ) (x, > x) (x,>x))
. x 1 ≤ x 2 , x 2 ≤ x 1 x1 \leq x2, x2 \leq x1 x1≤x2,x2≤x1 implies the solutions . . . ( − 1 , − 1 ) , ( 0 , 0 ) , ( 1 , 1 ) . . . ... (-1, -1), (0, 0), (1, 1) ... ...(−1,−1),(0,0),(1,1)... (abbr., ( x , x ) (x,x) (x,x))
. x 1 + 1 ≤ x 2 , x 2 + 1 ≤ x 1 x1 + 1\leq x2, x2 + 1\leq x1 x1+1≤x2,x2+1≤x1 implies no-solution, that is inconsistent.

As a lemma, if ( c 1 , c 2 , c 3 , . . . ) (c1, c2, c3, ...) (c1,c2,c3,...) is a solution, then ( c 1 + k , c 2 + k , c 3 + k , . . . ) (c1 + k, c2 + k, c3 + k, ...) (c1+k,c2+k,c3+k,...) is also a solution. (cuz x i + w ≤ x j xi + w \leq xj xi+w≤xj then x i + w + k ≤ x j + k xi + w + k \leq xj + k xi+w+k≤xj+k)

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Cuz x i < x j xi < xj xi<xj can transformed to x i + 1 ≤ x j xi + 1 \leq xj xi+1≤xj
x i > x j → x j + 1 ≤ x i xi > xj \quad \to \quad xj +1 \leq xi xi>xj→xj+1≤xi
x i = x j → x j ≤ x i , x i ≤ x j xi = xj \quad \to \quad xj \leq xi, xi \leq xj xi=xj→xj≤xi,xi≤xj
x i ≥ x j → x j ≤ x i xi \geq xj \quad \to \quad xj \leq xi xi≥xj→xj≤xi

Consequently, the general form of a relation is x i + c R x j xi + c \ R \ xj xi+c R xj where R R R can be = , < , > , ≤ , ≥ =, <, >, \leq, \geq =,<,>,≤,≥ and c c c is a constant.
. But, the relation x i R c xi \ R \ c xi R c where c c c is a constant is Invalid, it can’t be contained in this system, cuz it not satisfies the form x i + c ≤ x j xi +c \ \leq \ xj xi+c ≤ xj.
. Moreover, the relation x i + x j ≤ c xi + xj \leq c xi+xj≤c is invalid, it corresponds to x i − c ≤ − x j xi - c \leq -xj xi−c≤−xj, note this case.

选择 ≤ , ≥ \leq,\geq ≤,≥而不是 < , > <,> <,>

Firstly, let’s introduce the notion of Constant-Compare and Variable-Compare.
Constant-Compare: x < y x < y x<y where x , y x, y x,y are both certain constant (e.g., x = 1 , y = 2 x=1, y=2 x=1,y=2).
. let a , b , c a, b, c a,b,c be arbitrary constant, if we have a < b a < b a<b and b < c b < c b<c, then there must have a < c a < c a<c; more generally, ( a + w 1 < b , b + w 2 < c ) → ( a + w 1 + w 2 < c ) (a+w1<b, b+w2<c) \to (a+w1+w2<c) (a+w1<b,b+w2<c)→(a+w1+w2<c), cuz a + w 1 < b < c − w 2 a + w1 < b < c - w2 a+w1<b<c−w2, and then a + w 1 + w 2 < b + w 2 < c a+w1+w2<b+w2<c a+w1+w2<b+w2<c; we call it the property of Transitive. (i.e., we deduced a new relation based on two relations)
Variable-Compare: x < y x < y x<y where x , y x, y x,y are variable, that is the value of x , y x, y x,y has multiple choice.
. let x , y , z x, y, z x,y,z be variable, if we have x < y x < y x<y and y < z y < z y<z, then it not satisfies the property Transitivity. More precisely, x < z x < z x<z is just a necessity, but not a Necessity-Sufficiency. Cuz based on x < y , y < z x<y, y<z x<y,y<z we can deduced that z > x + 1 z > x + 1 z>x+1 (cuz the range of y y y is [ x + 1 , ∞ ) [x +1, \infty) [x+1,∞) from x < y x<y x<y, the range of z z z is [ y + 1 , ∞ ) [y + 1, \infty) [y+1,∞) from y < z y < z y<z, so combine them we conclude that the range of z z z is [ x + 2 , ∞ ) [x+2, \infty) [x+2,∞), so, x < z x < z x<z is just a necessity of x + 1 < z x + 1 < z x+1<z which is the correct-relation. (e.g., x < y , y < z , z < k x<y, y<z, z<k x<y,y<z,z<k, we have x + 2 < k x + 2<k x+2<k)
. This maybe some sophisticated, let’s discuss further, the reason why Transitivity must based on Necessity-Sufficiency is that, two relations R 1 , R 2 R1, R2 R1,R2 deduced a new relation R 3 R3 R3, so that R 3 R3 R3 can totally replace these previous two relations! that is R 1 : x + w 1 < y R1: x + w1 < y R1:x+w1<y and R 2 : y + w 2 < z R2: y + w2 < z R2:y+w2<z, then we know that z z z is confined by R 1 , R 2 R1, R2 R1,R2 simultaneously, now we want to use just one relation R 3 R3 R3 to replace the role acted above by R 1 , R 2 R1, R2 R1,R2. Therefore, R 3 R3 R3 must be equivalent to R 1 , R 2 R1, R2 R1,R2 (not just necessity). For example, x < y , y < z x<y, y<z x<y,y<z tells us that x + 1 < z x + 1 < z x+1<z; however the relation x < z x < z x<z deduced by Transitivity tells us x < z x < z x<z which indicates that z z z can choose x + 1 x + 1 x+1, but in fact z z z can’t be x + 1 x+1 x+1.

In summary, < , > <, > <,> has the property of Transitivity under Constant-Compare, but not for Variable-Compare.
That is, ( a 1 + w 1 < a 2 , a 2 + w 2 < a 3 , . . . , a m + w m < a k ) → ( a 1 + w 1 + w 2 + . . . + w m < a k (a1 + w1 < a2, a2 + w2 < a3, ..., am + wm < ak) \to (a1 + w1 + w2 + ... + wm < ak (a1+w1<a2,a2+w2<a3,...,am+wm<ak)→(a1+w1+w2+...+wm<ak satisfied only under Constant-Compare (i.e., a i ai ai is constant).

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Let’s see the case of ≤ , ≥ \leq, \geq ≤,≥.
You can prove that, the Transitivity ( a 1 + w 1 ≤ a 2 , a 2 + w 2 ≤ a 3 , . . . , a m + w m ≤ a k ) → ( a 1 + w 1 + w 2 + . . . + w m ≤ a k (a1 + w1 \leq a2, a2 + w2 \leq a3, ..., am + wm \leq ak) \to (a1 + w1 + w2 + ... + wm \leq ak (a1+w1≤a2,a2+w2≤a3,...,am+wm≤ak)→(a1+w1+w2+...+wm≤ak satisfied whichever Constant-Compare or Variable-Compare (i.e., it also satisfied if a i ai ai is variable).

Cuz this question is based on Variable-Compare and only ≤ ≥ \leq \geq ≤≥ satisfied the Transitivity, so we should transform all relation to the form x i + w 1 ≤ x j xi + w1 \leq xj xi+w1≤xj ( ≥ \geq ≥ is also usable, we will talk about it below, but the default choice is ≤ \leq ≤)

There is a rule (or stipulation), the constant-value w w w should always be placed on the left-side, that is the form x + w ≤ y x + w \leq y x+w≤y and x + w ≥ y x + w \geq y x+w≥y. If not, you should move the constant w w w to the left.

根据不等式组进行建图

We introduced Transitivity above, now we talk about its application.

For a relation x + w ≤ y x + w \leq y x+w≤y, cuz it shows a infinite solution set . . . ( − 1 , ≥ ( − 1 + w ) ) , ( 0 , ≥ ( 0 + w ) ) , ( 1 , ≥ ( 1 + w ) ) , . . . ... (-1, \geq (-1+w)), (0, \geq (0+w)), (1, \geq (1+w)), ... ...(−1,≥(−1+w)),(0,≥(0+w)),(1,≥(1+w)),..., or abbreviated ( x , ≥ ( x + w ) ) (x, \geq (x+w)) (x,≥(x+w)); we find that x x x has infinite choices . . . , − 1 , 0 , 1 , . . . ...,-1,0,1,... ...,−1,0,1,... and y y y has also infinite choices x + w , x + w + 1 , x + w + 2 , . . . x+w, x+w+1, x+w+2, ... x+w,x+w+1,x+w+2,..., that is two sides are both infinite. Our device is fixing the left-side, then only the right-side is infinite.
Subsequently, once x x x has fixed, then the solution/range of y y y is the form [ a , ∞ ) [a, \infty) [a,∞), therefore we can use a single number a a a to denote the minimal available number of y y y.

If we construct a graph by using a edge x → y x \to y x→y with weight w w w to represents the relation x + w ≤ y x + w \leq y x+w≤y.
e.g., for the fixed x x x, the relations x + 2 ≤ y x + 2 \leq y x+2≤y and x + 5 ≤ y x + 5 \leq y x+5≤y implies the range of y y y are [ x + 2 , . . . ) [x+2, ...) [x+2,...) and [ x + 5 , . . . ) [x+5, ...) [x+5,...) respectively, then the answer range of y y y should be intersection of these two ranges, that is [ x + 5 , . . . ) [x+5, ...) [x+5,...) is the answer, we use a single number x + 5 x+5 x+5 to denote the answer. Consequently, we should run the Greatest-Path on this graph.

Conclusion: We set the start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0, then run a Greatest-Path SPFA on G, if we found a Positive-Loop meaning that it’s No-Solution; otherwise, D i s t [ x ] Dist[x] Dist[x] means that the minimal-available-number of x x x under the condition D i s t [ a l l ] ≥ 0 Dist[all] \geq 0 Dist[all]≥0.

Cuz we have stipulated that x i xi xi is fixed and x j xj xj is confined by x i xi xi by the relation x i + w ≤ x j xi + w \leq xj xi+w≤xj; cuz x i xi xi maybe also confined by x z + w ≤ x i xz + w \leq xi xz+w≤xi, that is x j xj xj is confined by two relation, and maybe more;
So, any restriction on x i xi xi is a chain x a + w ≤ x b , x b + w ≤ x c , . . . , x d + w ≤ x i xa + w \leq xb, xb + w \leq xc, ..., xd + w \leq xi xa+w≤xb,xb+w≤xc,...,xd+w≤xi, according to the property of Transitivity, it can be combined into x z + w 1 + w 2 + . . . ≤ x i xz + w1 + w2 + ... \leq xi xz+w1+w2+...≤xi; we found that it corresponds to a path on the graph, i.e., any path s → . . . → x i s \to ... \to xi s→...→xi means a restriction on x i xi xi; e.g., two paths D i s t [ s 1 ] + w 1 = d 1 Dist[s1] + w1 = d1 Dist[s1]+w1=d1 and D i s t [ s 2 ] + w 2 = d 2 Dist[s2] + w2 =d2 Dist[s2]+w2=d2 means that the range of x i xi xi should be [ d 1 , ∞ ) [d1, \infty) [d1,∞) and [ d 2 , ∞ ) [d2, \infty) [d2,∞), and then we choose the intersection m a x ( d 1 , d 2 ) max(d1,d2) max(d1,d2) which corresponds to the Greatest-Path.

According to the property of Greatest-Path, if it is Solvable, D i s t [ a l l ] ≥ 0 Dist[all] \geq 0 Dist[all]≥0 finally. Cuz D i s t [ x ] Dist[x] Dist[x] is unique, the answer solution-set D i s t [ . . . ] Dist[...] Dist[...] is unique;
e.g., x + 1 ≤ y , y + 0 ≤ z , z + 0 ≤ y , x + 2 ≤ w x + 1 \leq y, y + 0 \leq z, z + 0 \leq y, x + 2 \leq w x+1≤y,y+0≤z,z+0≤y,x+2≤w, finally we get the solution-set: D i s t [ x ] = 0 , D i s t [ y , z ] = 1 , D i s t [ w ] = 2 Dist[x] = 0, Dist[y,z] = 1, Dist[w] = 2 Dist[x]=0,Dist[y,z]=1,Dist[w]=2 which is unique; in other words, under the condition D i s t [ a l l ] ≥ 0 Dist[all] \geq 0 Dist[all]≥0, the minimal value of x x x is 0 0 0 which is unique ( 0 0 0 maybe also the only available choice of x x x under this solution-set (i.e., ( x , 1 , 1 , 2 ) (x,1,1,2) (x,1,1,2) is not a solution-set if x ≠ 0 x \neq 0 x=0), but also maybe not (i.e., ( x , 1 , 1 , 2 ) (x,1,1,2) (x,1,1,2) is also a solution-set for some values x ≠ 0 x \neq 0 x=0), it will discussed further in the next section).
As a lemma, under the condition x i ≥ 0 xi \geq 0 xi≥0, two statements: 1. find out the minimal-value for every x i xi xi and 2. find out the minimal-value of the sum of all x i xi xi, are equivalent.

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x − 5 ≤ y x - 5 \leq y x−5≤y, when we set x = 0 x = 0 x=0, the range of y y y is [ − 5 , ∞ ) [-5, \infty) [−5,∞), but the algorithm will get D i s t [ y ] = 0 Dist[y] = 0 Dist[y]=0 not − 5 -5 −5; because we have the restriction The value of all variables should ≥ 0 \geq 0 ≥0.

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According to the property of Greatest-Path and set all-points be the start-points with the same D i s t [ a l l ] = c Dist[all]=c Dist[all]=c, the answer got above under the condition x i ≥ 0 xi \geq 0 xi≥0 can be generalized to the condition x i ≥ d xi \geq d xi≥d.
That is, set the start-points D i s t [ a l l ] = d Dist[all] = d Dist[all]=d not 0 0 0, we have D i s t [ x ] = D [ x ] + d Dist[x] = D[x] + d Dist[x]=D[x]+d where D [ x ] D[x] D[x] is the D i s t [ x ] Dist[x] Dist[x] under the case setting start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0.

≤ \leq ≤和 ≥ \geq ≥的等价性

Cuz the Inequality-Group is unchanged, x i + w ≤ x j xi + w \leq xj xi+w≤xj can also transformed to x j − w ≥ x i xj - w \geq xi xj−w≥xi;
The rules are the same as previous, assuming x j xj xj is fixed, then the range of x i xi xi is ( ∞ , x j − w ] (\infty, xj - w] (∞,xj−w] (this differs from previous), now the intersection is m i n ( d 1 , d 2 ) min(d1, d2) min(d1,d2) for two ranges ( ∞ , d 1 ] (\infty, d1] (∞,d1] and ( ∞ , d 2 ] (\infty, d2] (∞,d2], so it corresponds to Least-Path.

That is, transfer all relations to the form x i + c ≥ x j xi + c \geq xj xi+c≥xj, set the start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0, and then run Least-Path (If no Negative-Loop), we will get D i s t [ a l l ] ≤ 0 Dist[all] \leq 0 Dist[all]≤0; e.g., D i s t [ x ] = k Dist[x] = k Dist[x]=k means the maximal-available-value of x x x is k k k under the condition ∀ x i ≤ 0 \forall xi \leq 0 ∀xi≤0.

So, the maximal-value of x i xi xi under the condition x i ≤ d xi \leq d xi≤d equals D i s t [ x ] + d Dist[x] + d Dist[x]+d.

Moreover, there is a conclusion: D 1 [ x ] = − 1 ∗ D [ x ] D1[x] = -1 * D[x] D1[x]=−1∗D[x] where D [ x ] D[x] D[x] is the D i s t [ x ] Dist[x] Dist[x] (minimal-available-value) under the Greatest-Path of setting start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0 with the form x i + w ≤ x j xi + w \leq xj xi+w≤xj, and D 1 [ x ] D1[x] D1[x] (maximal-available-value) under the Least-Path of setting start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0 with the form x i + w ≥ x j xi + w \geq xj xi+w≥xj.

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Therefore, the case of ≥ \geq ≥ could also always transferred to ≤ \leq ≤.
For instance, get the maximal-value under the condition x i ≤ d xi \leq d xi≤d;
Firstly, we transform all relations to the form x i + w ≤ x j xi + w \leq xj xi+w≤xj, build a graph, set start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0, run a Greatest-Path SPFA; finally, D i s t [ x ] Dist[x] Dist[x] is the minimal-value of x x x under the condition ∀ x i ≥ 0 \forall xi \geq 0 ∀xi≥0
Secondly, − 1 ∗ D i s t [ x ] -1 * Dist[x] −1∗Dist[x] means the maximal-value of x x x under the condition ∀ x i ≤ 0 \forall xi \leq 0 ∀xi≤0;
Further, − 1 ∗ D i s t [ x ] + d -1 * Dist[x] + d −1∗Dist[x]+d is the maximal-value of x x x under the condition ∀ x i ≤ d \forall xi \leq d ∀xi≤d.

几何解释

If the condition is ∀ x i ≥ d \forall xi \geq d ∀xi≥d, it means that every number x i xi xi in the number-axis should as closely as possible to d d d from the right-side (i.e., x i ≥ d xi \geq d xi≥d) (the condition ∀ x i ≤ d \forall xi \leq d ∀xi≤d means from the left-side), so that the sum of ∣ x i − d ∣ |xi - d| ∣xi−d∣ attains the minimum among other solutions.

做题技巧

Suppose that there is a structure SPFA: Check-Positive-Loop with Start-Distance be 0 0 0 (that is, setting start-points D i s t [ a l l ] = 0 Dist[all] = 0 Dist[all]=0 and then run Greatest-Path SPFA), then all Inequality-Group problems can be solvable using this same structure.
In other words, if your device for Inequality-Group using a different SPFA which is not SPFA: Check-Positive-Loop with Start-Distance be 0 0 0, then, your device is wrong.

对恒等关系 x i = k xi = k xi=k的特殊处理

If there is relation x i = k xi = k xi=k, it can be also solvable using a trick by transfer it to x i = x j + c xi = xj + c xi=xj+c where c c c is also a constant, then for the relation x i = x j + c xi = xj + c xi=xj+c that is valid, it corresponds to two relations x i ≤ x j + c xi \leq xj + c xi≤xj+c and x i ≥ x j + c xi \geq xj + c xi≥xj+c which correspond to the standard relations x i − c ≤ x j xi - c \leq xj xi−c≤xj and x j + c ≤ x i xj + c \leq xi xj+c≤xi.

If you ensure that there always exists a point x j xj xj such that D i s t [ x j ] Dist[xj] Dist[xj] always equals to a constant c c c whatever the graph G (the premise is G is solvable), then x i = k = k + c − c = c + ( k − c ) = x j + ( k − c ) xi = k = k + c - c = c + (k - c) = xj + (k-c) xi=k=k+c−c=c+(k−c)=xj+(k−c), that is x i = x j + ( k − c ) xi = xj + (k-c) xi=xj+(k−c) where k , c k,c k,c are knowns, so it can transferred to x i − ( k − c ) ≤ x j xi - (k-c) \leq xj xi−(k−c)≤xj and x j + ( k − c ) ≤ x i xj + (k-c) \leq xi xj+(k−c)≤xi;

For example, if there is a Zero-Chain that involves all points x 1 → x 2 → . . . → x n x1 \to x2 \to ... \to xn x1→x2→...→xn with all edges be 0 0 0, there has a property from Greatest-Path: If G is solvable (no Positive-Loop), then x 1 x1 x1 must be a real-start-point, that is D i s t [ x 1 ] Dist[x1] Dist[x1] always equals 0 0 0. So, we conclude that D i s t [ x 1 ] = 0 Dist[x1] = 0 Dist[x1]=0 whatever the graph G is but contains a Zero-Chain and be solvable;
Therefore, a relation x i = c xi = c xi=c can be transformed to x i = c + 0 = x 1 + c xi = c + 0 = x1 + c xi=c+0=x1+c, then x i − c ≤ x 1 xi -c \leq x1 xi−c≤x1 and x 1 + c ≤ x i x1 + c \leq xi x1+c≤xi

转换为前缀和数组时的注意点

If you get a inequality x l + . . . + x r R c xl + ... + xr \ R \ c xl+...+xr R c, we find that if we transform it to Prefix-Sum-Array, then it corresponds to S [ r ] − S [ l − 1 ] R c S[r] - S[l-1] \ R \ c S[r]−S[l−1] R c which satisfies the standard inequality form.

Note that, the index of x i xi xi should starts at 1 1 1 so that every x i xi xi corresponds to two Prefix-Sum. (in other words, x 0... x n − 1 x0...x_{n-1} x0...xn−1​$ should move abstractly to x 1... x n x1...xn x1...xn, then we have a Prefix-Sum S 0 , . . . , S n S0, ..., Sn S0,...,Sn, that is S 1 − S 0 = x 0 S1 - S0 = x0 S1−S0=x0)
If you use S 0 = x 0 , S i − S i − 1 = x i S0 = x0, Si - S_{i-1} = xi S0=x0,Si−Si−1​=xi, then for a relation x l + . . . x r R c xl + ... xr \ R \ c xl+...xr R c where if l = 0 l = 0 l=0 then it corresponds to a relation with one variable S r R c Sr \ R \ c Sr R c, and corresponds to a relation with two variables S r − S l − 1 R c Sr - S_{l-1} \ R \ c Sr−Sl−1​ R c if l > 0 l > 0 l>0;
This makes a bit of cumbersome cuz you need handle by cases, and moreover, a relation with just one relation S r R c Sr \ R \ c Sr R c is not valid.

例题

1169. 糖果

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There are two types inequalities: x i ≥ 1 ∀ x i xi \geq 1 \ \ \forall xi xi≥1  ∀xi and x i R x j xi \ R \ xj xi R xj where R R R is = , < , > , ≤ , ≥ =,<,>,\leq,\geq =,<,>,≤,≥ (it can transformed to x i + w ≤ x j xi + w \leq xj xi+w≤xj). Then calculate the minimal-sum of x i xi xi.

It is the standard Inequality-Group form, so after the algorithm, you will get the solution-set for the minimal value of x i ≥ 0 xi \geq 0 xi≥0, then perform x i = x i + 1 xi = xi + 1 xi=xi+1 to making sure x i ≥ 1 xi \geq 1 xi≥1;

362. 区间

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A similar question link which says that the interval has o d d / e v e n odd/even odd/even number of 1 1 1 (Disjoint-Set with Depth can solve it)

Let A i Ai Ai denotes whether it is 1 1 1 in the position of i i i where the index of A A A starts at 1 1 1, then A i = 0 / 1 Ai = 0/1 Ai=0/1.
We have the relation:
A i ≥ 0 Ai \geq 0 Ai≥0
A i ≤ 1 Ai \leq 1 Ai≤1
A l + . . . + A r ≥ c Al + ... + Ar \geq c Al+...+Ar≥c (not satisfy the standard-form)

When involving a sum of interval elements, it usually transferred to Prefix-Sum (there also has a trick that there must exists a empty element in Prefix-Sum S 0 S0 S0; in other words, n n n variables x 1 , . . . , x n x1, ..., xn x1,...,xn, correspond to n + 1 n + 1 n+1 variables S 0 , S 1 , . . . , S n S0, S1, ..., Sn S0,S1,...,Sn when transformed to Prefix-Sum; the reason why doing so is explained in link)

Now
S i − S i − 1 ≥ 0 Si - S_{i-1} \geq 0 Si−Si−1​≥0
S i − S i − 1 ≤ 1 Si - S_{i-1} \leq 1 Si−Si−1​≤1
S r − S l − 1 ≥ c Sr - S_{l-1} \geq c Sr−Sl−1​≥c
which is a standard Inequality-Group.

The answer (the sum of A 1 , . . . , A n A1, ..., An A1,...,An) equals S n − S 0 Sn - S0 Sn−S0.

More speaking, when handling with Prefix-Sum, usually it will exists a Zero-Chain, as here S 0 → S 1... → S n S0 \to S1 ... \to Sn S0→S1...→Sn where every edge is 0 0 0, it means that if it is consistent, then S 0 S0 S0 must be a real-start-point, that is, D i s t [ S 0 ] Dist[S0] Dist[S0] must be 0 0 0 in any cases.

1170. 排队布局

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n n n points on a line from left to right x 1 ≤ . . . ≤ x n x1 \leq ... \leq xn x1≤...≤xn
It has the relations:
x i ≤ x i + 1 ∀ i ∈ [ 1 , n ) xi \leq x_{i+1} \quad \forall i \in [1, n) xi≤xi+1​∀i∈[1,n)
x j − x i ≤ c xj - xi \leq c xj−xi≤c where j > i j > i j>i
x j − x i ≥ c xj - xi \geq c xj−xi≥c where j > i j > i j>i
It already could be transformed to the standard-form (where has no condition of the x i ≤ o r ≥ d ∀ x i xi \leq or \geq d \quad \forall xi xi≤or≥d∀xi which is a optional restriction);

After the algorithm, if it returns False that is it is inconsistent; otherwise we got the minimal value D i s t [ x ] Dist[x] Dist[x] for every x x x such that x i ≥ 0 xi \geq 0 xi≥0.

How should we calculate the maximal-distance between x 1 , x n x1, xn x1,xn?
After the algorithm, we get D i s t [ x 1 ] = 0 Dist[x1] = 0 Dist[x1]=0 and D i s t [ x 1 ] ≤ D i s t [ x 2 ] ≤ . . . ≤ D i s t [ x n ] Dist[x1] \leq Dist[x2] \leq ... \leq Dist[xn] Dist[x1]≤Dist[x2]≤...≤Dist[xn] (due to the Zero-Chain x 1 → x 2 → . . . → x n x1 \to x2 \to ... \to xn x1→x2→...→xn), as we discussed in the article above, for this current solution-set a 1 , a 2 , . . . , a n a1, a2, ..., an a1,a2,...,an where a i = D i s t [ x i ] ai = Dist[xi] ai=Dist[xi], for any x i xi xi, there are two cases: 1 maybe the solution-set a 1 , . . . [ a i , ∞ ) , . . . , a n a1, ... [ai, \infty), ..., an a1,...[ai,∞),...,an is also valid under the condition of without modify a 1 a1 a1; 2 maybe just a few of solution-sets a 1 , . . . [ a i , a i + k ] , . . . , a n a1, ... [ai, ai + k], ..., an a1,...[ai,ai+k],...,an are valid under the condition of without modify a 1 a1 a1.
1 e.g., n = 3 n=3 n=3 with x 1 ≤ x 2 ≤ x 3 x1 \leq x2 \leq x3 x1≤x2≤x3 and x 2 ≥ x 3 x2 \geq x3 x2≥x3, we also got the solution-set ( 0 , 0 , 0 ) (0,0,0) (0,0,0), we find that 0 , x , x 0, x, x 0,x,x where x ∈ [ 0 , ∞ ) x \in [0, \infty) x∈[0,∞) is also a solution.
2 e.g., n = 3 n=3 n=3 with x 1 ≤ x 2 ≤ x 3 x1 \leq x2 \leq x3 x1≤x2≤x3 and x 2 ≥ x 3 , x 1 ≥ x 2 x2 \geq x3, x1 \geq x2 x2≥x3,x1≥x2, the solution-set is ( 0 , 0 , 0 ) (0,0,0) (0,0,0), however, the range of x n xn xn could only be [ 0 , 0 ] [0, 0] [0,0] not infinite (if x 3 = 1 x3 = 1 x3=1, then it must will affect x 1 x1 x1)
3 e.g., n = 3 n=3 n=3 with x 1 ≤ x 2 ≤ x 3 x1 \leq x2 \leq x3 x1≤x2≤x3 and x 2 ≥ x 3 , x 1 ≤ x 2 , x 2 − 2 ≤ x 1 x2 \geq x3, x1 \leq x2, x2 - 2 \leq x1 x2≥x3,x1≤x2,x2−2≤x1, the solution-set is ( 0 , 0 , 0 ) (0,0, 0) (0,0,0), but the solution-set ( 0 , 1 , 1 ) (0,1,1) (0,1,1) ( 0 , 2 , 2 ) (0,2,2) (0,2,2) are also solution-sets without modify x 1 x1 x1.

This is, according to D i s t [ ] Dist[] Dist[], we could not check that whether the maximal-distance between x 1 , x n x1, xn x1,xn is infinite or not.

Conclusion: The maximal-distance between x 1 , x n x1, xn x1,xn is infinite then there would not exists a path x n → . . . → x 1 xn \to ... \to x1 xn→...→x1; if exists such path, then the Greatest-Path D i s t [ x 1 ] Dist[x1] Dist[x1] with x n xn xn be start-point indicates the maximal-distance.
Proof: this path means a relation x n + w ≤ x 1 xn + w \leq x1 xn+w≤x1 and we already have a relation x 1 ≤ x n x1 \leq xn x1≤xn, therefore, x n − x 1 ≥ 0 xn - x1 \geq 0 xn−x1≥0 and x n − x 1 ≤ − w xn - x1 \leq -w xn−x1≤−w we found that the distance x n − x 1 xn - x1 xn−x1 has confined in − w -w −w (moreover, − w -w −w should ≥ 0 \geq 0 ≥0, otherwise it causes a contradiction which means inconsistent)

The Maximal-distance between x 1 , x n x1, xn x1,xn is that, we set D i s t [ x n ] = 0 Dist[xn] = 0 Dist[xn]=0 as the only start-point, then run Greatest-Path, finally D i s t [ 0 ] ∗ − 1 Dist[0] * -1 Dist[0]∗−1 is the answer.
Proof: cuz the graph G is unchanged, that is, a edge a → b a \to b a→b with w w w always denotes a relation a + w ≤ b a + w \leq b a+w≤b (that is no matter to whether the start-points is [ a l l ] = 0 [all] = 0 [all]=0 or [ x n ] = 0 [xn]=0 [xn]=0); e.g., there are two paths d 1 ≤ x i d1 \leq xi d1≤xi and d 2 ≤ x i d2 \leq xi d2≤xi then their intersection is m a x ( d 1 , d 2 ) max(d1, d2) max(d1,d2); in other words, the properties satisfies as previous are also valid when the start-point is just x n xn xn; so, it is still Greatest-Path.
Cuz the Zero-Chain (suppose that x 1 x1 x1 is accessible for x n xn xn), so we still the property D i s t [ x 1 ] ≤ . . . ≤ D i s t [ x n ] Dist[x1] \leq ... \leq Dist[xn] Dist[x1]≤...≤Dist[xn], and x n xn xn must be a real-start-point due to it it the only start-point and G is solvable, that is D i s t [ x n ] Dist[xn] Dist[xn] always equals 0 0 0;
Any path x n → . . . → x 1 xn \to ... \to x1 xn→...→x1 actually means a relation x n + w ≤ x 1 xn + w \leq x1 xn+w≤x1 (i.e., x n − x 1 ≤ − w xn - x1 \leq -w xn−x1≤−w, where w ≤ 0 w \leq 0 w≤0 proved above) which has already shows that the maximum of x n − x 1 xn - x1 xn−x1 is w w w; but the distance of x 1 , x n x1, xn x1,xn that is ∣ x 1 − x n ∣ = m a x ( x n − x 1 , x 1 − x n ) |x1 - xn| = max( xn - x1, x1 - xn) ∣x1−xn∣=max(xn−x1,x1−xn), now we got x n − x 1 xn - x1 xn−x1; we need consider x 1 − x n x1 - xn x1−xn; this should be linked to the previous information x 1 ≤ x n x1 \leq xn x1≤xn, so we have x 1 − x n ≤ 0 x1 - xn \leq 0 x1−xn≤0; therefore ∣ x 1 − x n ∣ = x n − x 1 |x1 - xn| = xn - x1 ∣x1−xn∣=xn−x1;
One path means the x n − x 1 ≤ − w xn - x1 \leq -w xn−x1≤−w (the maximum-distance is − w -w −w), but there maybe multiple paths, that is we have x n − x 1 ≤ − w 1 xn - x1 \leq -w1 xn−x1≤−w1, x n − x 1 ≤ − w 2 xn - x1 \leq -w2 xn−x1≤−w2, . . . ... ..., (it also represents x 1 ≥ w 1 , x 1 ≥ w 2 , . . . x1 \geq w1, x1 \geq w2, ... x1≥w1,x1≥w2,...), according to the intersection x 1 ≥ m a x ( w 1 , w 2 , . . . ) x1 \geq max(w1, w2, ...) x1≥max(w1,w2,...), and also the Greatest-Path would get m a x ( w 1 , w 2 , . . . ) max(w1, w2, ...) max(w1,w2,...) finally; so finally D i s t [ x 1 ] = k Dist[x1] = k Dist[x1]=k means that the relations between x 1 , x n x1, xn x1,xn must satisfies x n − x 1 ≤ − k xn - x1 \leq -k xn−x1≤−k (that is, if x n xn xn is fixed at 0 0 0, then the longest position which can be chosen by x 1 x1 x1 is − k -k −k);

One difference from the previous is that, in the previous D i s t [ x ] = m a x ( 0 , a l l p a t h s ) Dist[x] = max( 0, all \ paths) Dist[x]=max(0,all paths) due to every points is a start-point; but in here D i s t [ x ] = m a x ( a l l p a t h s ) Dist[x] = max( all \ paths) Dist[x]=max(all paths) due to there is only one start-point

Minimize_InequalitiesSystem< int, int> check( &G);
check.Initialize();
auto ret = check.Work();
if( false == ret){ //< inconsistentcout<< -1;return;
}
check.Spfa(); //< set `Dist[xn]=0` with others be `invalid`, run Grestest-Path
int invalid;
memset( &invalid, -0x7F, sizeof( invalid));
if( spfa.Distance[ 0] == invalid){ //< not accessiblecout<< -2;return;
}
cout<< -1 * spfa.Distance[ 0];

393. 雇佣收银员

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There are some terms:
we divide one-day into 24 24 24 hour-interval, [ 0 , 1 ] , [ 1 , 2 ] , . . . , [ 23 , 0 ] [0,1], [1,2], ..., [23,0] [0,1],[1,2],...,[23,0]; a staff has a start-work-time x x x, means this staff would work with 8 8 8 hour-intervals [ x , x + 1 ] , [ x + 1 , x + 2 ] , . . . , [ x + 7 , x + 8 ] [x,x+1], [x+1,x+2], ..., [x+7, x+8] [x,x+1],[x+1,x+2],...,[x+7,x+8];
M i [ x ] Mi[x] Mi[x] denotes there must have at least M i [ x ] Mi[x] Mi[x] staffs at the time [ x , x + 1 ] [x,x+1] [x,x+1]
M a [ x ] Ma[x] Ma[x] denotes there are M a [ x ] Ma[x] Ma[x] staffs whose start-work-time is x x x (that is, you can choose at most M a [ x ] Ma[x] Ma[x] at this time)

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Let’s see a wrong device using Greedy.
Let A [ x ] A[x] A[x] be the total staff-count at the time [ x , x + 1 ] [x, x+1] [x,x+1], then we have A [ x ] ≥ M i [ x ] A[x] \geq Mi[x] A[x]≥Mi[x], these A [ x ] A[x] A[x] staffs consists of a a a staffs whose start-work-time is x x x and b b b staffs whose start-work-time is < x < x <x.
If there exists a point x x x such that b = 0 b = 0 b=0 (i.e., A [ x ] = a A[x] = a A[x]=a) and A [ x ] = M i [ x ] A[x] = Mi[x] A[x]=Mi[x], let’s stop this idea cuz if we thinking rigorously you will find this is impossible, due to this problem is a Loop not linear, so there would not exists a Start-Point.
Suppose we choose the staffs be 0 , 7 , 14 , 21 0, 7, 14, 21 0,7,14,21, then A [ ] = [ 2 , 2 , 2 , 2 , 2 , 1 , 1 , 2 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 1 , 1 , 1 , . . . ] A[] = [2,2,2,2,2,1,1,2,1,1,1, 1,1,1,2,1,1,1,...] A[]=[2,2,2,2,2,1,1,2,1,1,1,1,1,1,2,1,1,1,...] and we let M i [ ] = A [ ] , M a [ ] > A [ ] Mi[] = A[], Ma[] > A[] Mi[]=A[],Ma[]>A[], we found that there is not a point x x x such that A [ x ] A[x] A[x] equals the staff-chosen-count whose start-time is x x x; ( A [ 0 / 7 / 14 / 21 ] = 2 A[0/7/14/21] = 2 A[0/7/14/21]=2 while the staff-chosen-count at that time is 1 1 1; other A [ ] > = 1 A[] >= 1 A[]>=1 while the staff-chosen-count is 0 0 0)

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Let’s see a wrong device using Inequality.
If we define A [ x ] A[x] A[x] be the total staff-count at the time [ x , x + 1 ] [x, x+1] [x,x+1] (e.g., if there is a staff whose start-work-time is a a a, then it will cause A [ a ] , . . . , A [ a + 7 ] A[a], ..., A[a+7] A[a],...,A[a+7] added by 1 1 1.
Then we would have the next inequalities:
A [ x ] ≤ 0 A[x] \leq 0 A[x]≤0, A [ x ] ≥ M i [ x ] A[x] \geq Mi[x] A[x]≥Mi[x], A [ x ] ≤ M a [ x ] A[x] \leq Ma[x] A[x]≤Ma[x]
It is very important to realize that the relation A [ x ] ≤ M a [ x ] A[x] \leq Ma[x] A[x]≤Ma[x] is wrong, cuz A [ x ] A[x] A[x] is the working-staff-count at time x x x, while M a [ x ] Ma[x] Ma[x] is the maximal-staff-count whose start-work-time is x x x. (e.g., M a [ i ] = 1 , M a [ i + 1 ] = 0 Ma[i] = 1, Ma[i+1]=0 Ma[i]=1,Ma[i+1]=0, we choose a staff at i i i, but we have A [ i ] , A [ i + 1 ] , . . . , A [ i + 7 ] = 1 A[i],A[i+1], ..., A[i+7] = 1 A[i],A[i+1],...,A[i+7]=1, therefore A [ i + 1 ] > M a [ i + 1 ] A[i+1] > Ma[i+1] A[i+1]>Ma[i+1]. In fact, this is reasonable)
Actually, it should be corrected by A [ x ] ≤ M a [ x ] + M a [ x − 1 ] + . . . + M a [ x − 7 ] A[x] \leq Ma[x] + Ma[x-1] + ... + Ma[x-7] A[x]≤Ma[x]+Ma[x−1]+...+Ma[x−7]; it still is complicated due to the definition of A [ x ] A[x] A[x]; suppose that now we got all A [ 0 , 1 , 2... , 23 ] A[0,1,2...,23] A[0,1,2...,23], we still don’t know how many staffs chosen which is the answer asked. It is important to associate your algorithm to the answer that problem-asked.

---

Cuz the answer that question asked is the total-staff-chosen-count, so we define A [ x ] A[x] A[x] be the staff-chosen-count whose start-work-time is x x x; this definition is very vital, then the answer is clear that equals the sum A [ ] A[] A[].
Then we have:
A [ x ] ≥ 0 ∀ x ∈ [ 0 , 23 ] A[x] \geq 0 \quad \forall x \in [0, 23] A[x]≥0∀x∈[0,23]
A [ x ] ≤ M a [ x ] A[x] \leq Ma[x] A[x]≤Ma[x]
∑ i = x − 7 x A [ i ] ≥ M i [ x ] \sum_{i = x - 7}^{x} A[i] \geq Mi[x] ∑i=x−7x​A[i]≥Mi[x] (note here, only the staffs whose start-time is x − 7 , . . . , x x-7, ..., x x−7,...,x would cover the current-time [ x , x + 1 ] [x, x+1] [x,x+1])

Cuz involves the summation, so we need transform to Prefix-Sum; (as we mentioned above, Prefix-Sum must has a empty-element S [ 0 ] S[0] S[0]; so A [ 0 − 23 ] → S [ 0 − 24 ] A[0-23] \to S[0-24] A[0−23]→S[0−24])
Let A [ x ] = S [ x + 1 ] − S [ x ] A[x] = S[x + 1] - S[ x] A[x]=S[x+1]−S[x] (note that, S [ x ] S[x] S[x] is Linear that is it equals the sum A [ 0 ] + . . . + A [ x − 1 ] A[0]+...+A[x-1] A[0]+...+A[x−1]; although this problem is not Linear but Loop)
So we have
S [ x + 1 ] − S [ x ] ≥ 0 S[x+1] - S[x] \geq 0 S[x+1]−S[x]≥0
S [ x + 1 ] − S [ x ] ≤ M a [ x ] S[x+1] - S[x] \leq Ma[x] S[x+1]−S[x]≤Ma[x]
@ T a g 0 @Tag0 @Tag0

@Tag0 is very vital, cuz A [ ] A[] A[] is a Loop, for the sum ∑ i = x − 7 x A [ i ] \sum_{i = x - 7}^{x} A[i] ∑i=x−7x​A[i], when x < 7 x < 7 x<7 it will involves A [ 23 ] , A [ 22 ] . . . A[23], A[22]... A[23],A[22]...; while the Prefix-Sum S [ ] S[] S[] is Linear that is its Start-Point is A [ 0 ] A[0] A[0]; so the transformation here is vital;

So, it becomes:
S [ x + 1 ] − S [ x ] ≥ 0 S[x+1] - S[x] \geq 0 S[x+1]−S[x]≥0
S [ x + 1 ] − S [ x ] ≤ M a [ x ] S[x+1] - S[x] \leq Ma[x] S[x+1]−S[x]≤Ma[x]
S [ x + 1 ] − S [ x − 7 ] ≥ M i [ x ] i f x ≥ 7 S[x+1] - S[x - 7] \geq Mi[x] \quad if \ x \geq 7 S[x+1]−S[x−7]≥Mi[x]if x≥7
S [ x + 1 ] + ( S [ 24 ] − S [ 17 + x ] ) ≥ M i [ x ] i f x < 7 S[x+1] + (S[24] - S[17 + x]) \geq Mi[x] \quad if \ x < 7 S[x+1]+(S[24]−S[17+x])≥Mi[x]if x<7

The former 3 3 3 relations fit the standard-form x i + c ≤ x j xi + c \leq xj xi+c≤xj, while there exists 3 3 3 unknowns S [ x + 1 ] , S [ 24 ] , S [ 17 + x ] S[x+1], S[24], S[17+x] S[x+1],S[24],S[17+x] in the last relation, we need to eliminate it to 2 2 2 unknowns
Cuz for all inequalities with the last-form, they share the same unknown S [ 24 ] S[24] S[24].

So we can iterate it to make it a constant (that is, we try multiple Inequality-Systems), suppose S [ 24 ] = c S[24] = c S[24]=c, now although that inequality fit the standard-form, it is very important that now you have a new relation S [ 24 ] = c S[24] = c S[24]=c.

According to the trick mentioned above, cuz there is a Zero-Chain S 0 → . . . → S 24 S0 \to ... \to S24 S0→...→S24, so it is solvable, then S 0 S0 S0 must be 0 0 0, so S 24 = c = c + 0 = c + S 0 S24 = c = c + 0 = c + S0 S24=c=c+0=c+S0, then S 24 ≤ c + s 0 S24 \leq c + s0 S24≤c+s0, S 24 ≥ c + S 0 S24 \geq c + S0 S24≥c+S0;

---

We find that S 24 = c S24 = c S24=c denotes the answer ( S 24 S24 S24 is the sum of all A [ ] A[] A[]), and cuz the answer is the least number; so c c c satisfies Binary-Search.

bool Check( int _mid){for( int i = 0; i < 7; ++i){G->Modify_edge( Edges_0.at( i), Mi[ i] - _mid);}G->Modify_edge( Edges_1.at( 0), _mid);G->Modify_edge( Edges_1.at( 1), -1 * _mid);if( false == Spfa->Work()){return true;}return false;
}
//----for( int i = 0; i < 24; ++i){//> A[ i] >= 0 `->` Sum[ i + 1] - S[ i] >= 0G->Add_edge( i, i + 1, 0);
}
for( int i = 0; i < 24; ++i){//> A[ i] <= Ma[ i] `->` Sum[ i + 1] - S[ i] <= Ma[ i]G->Add_edge( i + 1, i, -1 * Ma[ i]);
}
for( int i = 7; i < 24; ++i){//> A[ i - 7] + ... + A[ i] >= Mi[ i] `->` S[ i + 1] - S[ i - 7] >= Mi[ i]G->Add_edge( i - 7, i + 1, Mi[ i]);
}
Edges_0.clear();
for( int i = 0; i < 7; ++i){//> A[ 23 - ?] + ... + A[ 23] + A[ 0] + ... + A[ i] >= Mi[ i] `->` S[ i + 1] + S[ 24] - S[ 17 + i] >= Mi[ i]G->Add_edge( 17 + i, i + 1, 0); //< `0` is just a placeholderEdges_0.push_back( G->Get_last_edgeId());
}
Edges_1.clear();
//> S[ 24] = C `->` S[ 24] >= S[ 0] + C, S[ 24] <= S[ 0] + C
G->Add_edge( 0, 24, 0);
Edges_1.push_back( G->Get_last_edgeId());
G->Add_edge( 24, 0, 0);
Edges_1.push_back( G->Get_last_edgeId());
//--
Spfa = new SPFA_CheckLoop< int, int>( G);
Spfa->Initialize();
int l = 0, r = n;
while( l < r){int mid = (l + r) >> 1;if( Check( mid)){r = mid;}else{l = mid + 1;}
}
if( false == Check( r)){cout<< "No Solution" <<endl;return;
}
cout<< r <<endl;

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