`Computer-Algorithm` 有向无环图DAG
Contents
- 有向无环图
- 定义
- 性质
- 应用
- 最小路径点覆盖
- 最小路径重复点覆盖
- 路径独立集
- 至少新添加几条边,使得DAG成为SCC
有向无环图
定义
性质
+ let HHH be the set of start-points and TTT the set of end-points in a Weakly-Connected-DAG;
. Any point (including H/TH/TH/T) must be reachable by at least one point of HHH and must can reach to at least one point of TTT;
. Any point a∈Ha \in Ha∈H and any point b∈Tb \in Tb∈T, aaa may not reach to bbb; e.g., (a→b)(a→c)(d→c)(a\to b) (a\to c) (d\to c)(a→b)(a→c)(d→c) where the start-point ddd cannot reach to the end-point bbb;
+ The 最大路径独立集 (it is a set of points such that any two points of it are not accessible to each other) of a DAG, maybe not the H/TH/TH/T where HHH is the set of start-points and TTT the set of end-points;
. e.g., (a→b)(a→c)(b→d)(c→d)(a\to b) (a \to c) (b\to d) (c\to d)(a→b)(a→c)(b→d)(c→d), the 最大路径独立集 is (b,c)(b,c)(b,c)
+ The transitive-graph of a DAG is also a DAG; (i.e., let E[a][b]E[a][b]E[a][b] denotes an edge in the DAG, then performing Floyd-Boolean to get a new graph GGG, finally GGG would also be a DAG)
+ Any path is DAG is Simple-Path (i.e., all points on the path are distinct);
+ The Source-Points of a DAG is the set of all points xxx such that din[x]=0din[ x] = 0din[x]=0, also the End-Points is all points dout[x]=0dout[x] = 0dout[x]=0;
. Any non-empty DAG must contains at least one Source-Points and End-Points. (let n,mn,mn,m be the number of Sources and Ends, then either n>0,m>0n>0, m>0n>0,m>0 or n=m=0n=m=0n=m=0, it is impossible that n=0,m>0n=0, m>0n=0,m>0)
. A point xxx is both Source and End if and only if xxx is a Isolated-Point.
+ Any two points a,ba,ba,b in a DAG maybe not Weakly-Connected (e.g., (1→2)(3)(1\to 2) (3)(1→2)(3) is a DAG); more precise, a,ba,ba,b either be Weakly-Connected, or Non-Connected; another phrase: a,ba,ba,b must be Non-Strongly-Connected.
+ (DAG can does Topological-Sorting) is equivalent with (Any graph that can does Topological-Sorting, must is a DAG)
. The Topological-Sequence of a DAG maybe not unique.
+ For any graph G=(V,E)G =(V,E)G=(V,E), after the performance Shrinking-SCC, it would becomes a DAG.
. Shrinking-SCC: let Id[v]Id[v]Id[v] denotes the SCC-ID of every point v∈Vv \in Vv∈V; Let G1=(V1,E1)G1 = (V1, E1)G1=(V1,E1) be a empty graph, and then iterating all edges (a→b)∈E(a \to b) \in E(a→b)∈E, add the edge Id[a]→Id[b]Id[a] \to Id[b]Id[a]→Id[b] into G1G1G1 when Id[a]≠Id[b]Id[a] \neq Id[b]Id[a]=Id[b]. Finally, G1G1G1 is a DAG.
. Moreover, G1G1G1 has no Loop, but may contains Multiple-Edge (you can use Hashing to avoid it when adding edges into G1G1G1);
For a Weakly-Connected DAG, let its Source be s1,s2,...,sns1,s2,...,sns1,s2,...,sn and its End be e1,e2,...,eme1,e2,...,eme1,e2,...,em;
We build a Bipartite-Graph where L=(s1,s2,...,sn),R=(e1,e2...,em)L = (s1,s2,..., sn), R = (e1,e2...,em)L=(s1,s2,...,sn),R=(e1,e2...,em); if sisisi can reach to ejejej, then we connected a edge between si,ejsi, ejsi,ej in the Bipartite;
As a result, the Maximal-Match of this Bipartite is min(n,m)min( n, m)min(n,m) (i.e., Full-Matched)
. Cuz any two points si,ejsi, ejsi,ej exists a path (i.e., if n<mn < mn<m, sisisi always has a Augmented-Path)
But if the DAG is not Weakly-Connected, it is not correct; e.g., if the DAG contains two Weakly-Connected-Sub-Graphs that are (2,3)(3,2)(2,3) (3,2)(2,3)(3,2) where (2,3)(2,3)(2,3) represents 222 Sources and 333 Ends;
The Maximal-Match is min(2,3)+min(3,2)=4min(2, 3) + min(3,2) = 4min(2,3)+min(3,2)=4, not min(2+3,3+2)=5min(2+3, 3+2) = 5min(2+3,3+2)=5
应用
最小路径点覆盖
If the graph is a DAG, 最小路径点覆盖 can be solved by Bipartite;
Let DDD be the DAG where all points are a1,a2,...,ana1,a2,...,ana1,a2,...,an, now we construct a Bipartite BBB using below method
for( (ai -> aj) : all edges in D){B.add( ai, aj +|V|);
}
In BBB, the L-Set are a1,...,ana1,...,ana1,...,an and R-Set are a1‾,...,an‾\overline{a1},...,\overline{an}a1,...,an where ai‾=a1+∣V∣\overline{ai} = a1 + |V|ai=a1+∣V∣ (no specific meaning, just making all ai‾\overline{ai}ai differs to aiaiai in the Bipartite); (if D is not DAG, B is also a Bipartite)
(最小路径点覆盖) equals ∣V∣−m|V| - m∣V∣−m where mmm is the Maximal-Match of BBB;
@Delimiter
证明
+ Cuz we only care about ∣S∣|S|∣S∣ where SSS is the 最小路径点覆盖, e.g., S=(a,b,c)(d,e)...S = (a,b,c) (d,e) ...S=(a,b,c)(d,e)... where every (...)(...)(...) denotes a path;
. We stipulate that a path (a,b,c)∈S(a,b,c) \in S(a,b,c)∈S actually denotes all paths with the form a→b→ca \to b \to ca→b→c (due to multiple edges between two points, so (a,b,c)(a,b,c)(a,b,c) may denotes many paths), it would not affect the calculation; cuz we focus on the number of ()()()() () ()()()(), not care about (?)(?)(?).
+ (路径点覆盖) →\to→ (Matches in Bipartite)
. (a1,a2,...ai)(ax,...)(..)...(a1,a2, ... ai) (ax,...) (..) ...(a1,a2,...ai)(ax,...)(..)... is a 路径点覆盖, where the points in all paths are distinct, any directed edge ai→ajai \to ajai→aj in any path had been transformed into undirected edge (ai−aj‾)(ai - \overline{aj})(ai−aj) in BBB; we found that all such edges in BBB are disjoints, that is all these edges in BBB form a Match;
. e.g., a 路径点覆盖 (a,b,c)(d,e)(f)(a, b, c) (d, e) (f)(a,b,c)(d,e)(f) in DAG corresponds to a Match (a−b‾)(b−c‾)(d−e‾)(a-\overline{b}) (b-\overline{c}) (d- \overline{e})(a−b)(b−c)(d−e) in DDD;
. This map is Injective; (if D is not DAG, this is also satisfies)
+ (Matches in Bipartite) →\to→ (路径点覆盖)
. For a match (ai−aj‾)()()...(ai-\overline{aj}) () () ...(ai−aj)()()..., cuz every edge (a−b‾)(a-\overline{b})(a−b) actually is a directed edge a→ba \to ba→b in DAG, from this view-point, firstly we write the match in the form (ai,aj‾)(xx,xx)...(ai,\overline{aj}) (xx,xx) ...(ai,aj)(xx,xx)... where we call a (xx)(xx)(xx) be a path (corresponds to a Simple-Path in DAG); if there exists (a,b‾)(c,d‾)(a,\overline{b}) (c, \overline{d})(a,b)(c,d) where b=cb=cb=c, then merge them to be (a,b,d‾)(a, b, \overline{d})(a,b,d) (cuz two such edges a→b,b→ca\to b, b \to ca→b,b→c corresponds to a path a→b→ca\to b \to ca→b→c); proceeding this process, finally, every path is in the form (a1,a2,...,ak‾)(a1, a2, ..., \overline{ak})(a1,a2,...,ak) (a1≠aka1 \neq aka1=ak, cuz DAG has no circuit);
. All paths are distinct, that is there is no point belongs to multiple paths; Proof: (a1,a2,a3‾)(a4,a5,a6‾)(a1,a2,\overline{a3}) (a4,a5,\overline{a6})(a1,a2,a3)(a4,a5,a6) all a1,a2,a4,a5a1, a2, a4, a5a1,a2,a4,a5 (all points except the last-points) are distinct (due to they come from the left-set of original matches); all a2,a3‾,a5,a6‾a2, \overline{a3}, a5, \overline{a6}a2,a3,a5,a6 (all points except the first-points) are distinct (due to they also come from the right-set of the original matches); ai,aj‾ai, \overline{aj}ai,aj (aiaiai is the first-point in a path, aj‾\overline{aj}aj is the last-point in a path) are distinct (i.e., ai≠ajai \neq ajai=aj) (cuz the merge-process has finished, they must be in the same path, then it indicates that the path is a Circuit which is impossible is DAG)
. Finally, add all Isolated-Points; e.g., the match corresponds to paths (a1,a2,a3‾)(a4,a5‾)(a1,a2,\overline{a3}) (a4, \overline{a5})(a1,a2,a3)(a4,a5) (there are 777 points in total), then add the isolated points a6,a7a6, a7a6,a7; the new paths are (a1,a2,a3)(a4,a5)(a6)(a7)(a1,a2,a3) (a4,a5) (a6) (a7)(a1,a2,a3)(a4,a5)(a6)(a7) which is a 路径点覆盖
. Different matches correspond to different 路径点覆盖; so the map is Injective;
As a result, (路径点覆盖) and (matches) forms a Bijection;
Now, let’s consider the paths-count of a Match; all paths in a Match has two kinds (as mentioned above): 1 a self-path, i.e., a isolated-point (a)(a)(a), in this case, aaa must be a Unmatched-Point in L-Set under the match; 2 a path (a1,...,ai‾)(a1,...,\overline{ai})(a1,...,ai), then aiaiai (note that, not ai‾\overline{ai}ai must be a Unmatched-Point in L-Set under the match (if not (that is, it is a matched-point of L-Set), then it would be a non-last-point in a path, which would contradicts with the context ai‾\overline{ai}ai is a last-point);
. So, every path corresponds to a Unmatched-Point in L-Set; cuz these paths contains all points, so there would not exist a Unmatched-Point in L-Set with which no path corresponds;
. In a match, the paths-count equals the Unmatched-Points in L-Set, that is n−Mn - Mn−M where MMM is the match-count;
Therefore, 最小路径点覆盖 equals n−mn - mn−m where mmm is the maximal-match;
最小路径重复点覆盖
If the graph DDD is a DAG, (最小路径重复点覆盖) equals (最小路径点覆盖) in DDDDDD; where DDDDDD is the Transitivity-Graph of DDD (i.e., two edges a,ba,ba,b and b,cb,cb,c in DDD produce a new edge a,ca,ca,c)
@Delimiter
证明
A same stipulation as above, we use (a,b,c)(a,b,c)(a,b,c) denotes any path in the form a→b→ca\to b \to ca→b→c due to the multiple edges between two points; it would not affect, cuz we only focus on the paths-count, not the edges of a path;
If the graph DDD is a DAG, for a (路径重复点覆盖), we transform it to a Simplified-Form: if a point aaa occurs in two paths ...a......a......a... and ...a......a......a..., let’s focus on the first path 1 if aaa is the first-point or last-point (a,...a, ...a,... or ...,a... , a...,a) then remove this point from the path, all points are still coved; 2 if aaa is ...x,a,y...... x , a , y ......x,a,y..., then we use x,yx , yx,y to replace x,a,yx , a , yx,a,y (note that x,yx , yx,y no longer means a edge in DDD (cuz DDD may have no edge between them), it is just a transformation), now all points are still covered;
Finally, these paths cover all all points only once (that is, becomes a 路径点覆盖); but as discussed above, it is not a 路径点覆盖 of DDD (a path ...,x,y,......, x,y, ......,x,y,... in 路径点覆盖 where x,yx,yx,y must denotes a edge in a DAG, while there maybe no edge x,yx,yx,y in DDD);
So, we need to add such edges x,yx,yx,y into DDD (i.e., run Floyd-Boolean, two edges a,ba,ba,b and b,cb,cb,c produces a new edge a,ca,ca,c); the new graph would still be a DAG; let the new DAG be DDDDDD;
It can be proved that, 路径重复点覆盖 in DDD is equivalent to 路径重复点覆盖 in DDDDDD (cuz if a path uses an edge x,yx,yx,y in DDDDDD, then it can be decomposed into x,...,yx,...,yx,...,y where all edges are raw-edges)
As a result, any 路径重复点覆盖 in DDDDDD, using the above Simplified-Form operation, it can be transformed to a 路径点覆盖 in DDDDDD; and all 路径点覆盖 are also 路径重复点覆盖;
so we just calculate the 最小路径点覆盖 in DDDDDD which equals to the 最小路径重复点覆盖 in DDDDDD (also in DDD);
@Delimiter
例题
AcWing-379. 捉迷藏
路径独立集
If DDD is a DAG, (最大独立集) equals (最小路径重复点覆盖) in DDD;
@Delimiter
证明
@Unfinished
至少新添加几条边,使得DAG成为SCC
AcWing-367. 学校网络
For a DAG (V,E)(V, E)(V,E), let xxx denoted the answer;
1 x=0x = 0x=0 if ∣V∣=1|V| = 1∣V∣=1
2 otherwise x=max(s,e)x = max(s, e)x=max(s,e) where sss is the number of Source-Points and eee is the End-Points;
For a DAG GGG, let its Source be s1,s2,...,sns1,s2,...,sns1,s2,...,sn and its End be e1,e2,...,eme1,e2,...,eme1,e2,...,em; you can imagine that a graph BBB which consists of these points, and if sisisi can reach to ejejej then BBB has a edge si→ejsi \to ejsi→ej; BBB is also a DAG, and once you make BBB a SCC then GGG is also a SCC.
Proof: for any points a,ba,ba,b in GGG and suppose that s1→a→e1s1 \to a \to e1s1→a→e1 and s2→b→e2s2 \to b \to e2s2→b→e2 where →\to→ means can reach; cuz s1,s2,e1,e2s1,s2,e1,e2s1,s2,e1,e2 are Strongly-Connected, there exists two paths a→e1→s2→ba \to e1 \to s2 \to ba→e1→s2→b and b→e2→s1→ab \to e2 \to s1 \to ab→e2→s1→a, therefore a,ba,ba,b are Strongly-Connected;
A easy device is adding these edges si→si−1si \to s_{i-1}si→si−1 and ei→ei+1ei \to e_{i+1}ei→ei+1 and em→s1em \to s1em→s1, the answer is (n−1)+(m−1)+1(n-1) + (m-1) + 1(n−1)+(m−1)+1, but it is not the optimal choice;
The correct approach involves the notion of Bipartite-Graph as mentions above.
A DAG GGG, corresponds to a Bipartite-Graph BBB where L=(s1,s2,...,sn),R=(e1,...,em)L = (s1,s2,...,sn), R = (e1,...,em)L=(s1,s2,...,sn),R=(e1,...,em) where LLL is the set of points whose din=0din = 0din=0 and RRR for dout=0dout = 0dout=0, if sisisi can reach to ejejej then there is a edge si→ejsi \to ejsi→ej in BBB; It is obvious that BBB is also a DAG; There is a very crucial property on of this Partite: any point must has a edge (there exists a edge si→?si \to ?si→? or ?→ei? \to ei?→ei, ∀si,ej\forall si, ej∀si,ej)
Our goal is to make BBB becomes a SCC;
The whole process is:
Originally, GGG is DAG with BBB as its Bipartite, so we have ∣L∣>0,∣R∣>0|L| >0, |R| > 0∣L∣>0,∣R∣>0;
1 If min(∣L∣,∣R∣)≥2min( |L|, |R| ) \geq 2min(∣L∣,∣R∣)≥2, then there must exists at least two matches on BBB;
. Disproof: let l1,l2,r1,r2l1,l2, r1,r2l1,l2,r1,r2, if the maximal-match is 111, e.g., the only match is l1−r1l1-r1l1−r1, then all li→r1li \to r1li→r1 (if not, li−rili - rili−ri is another match); let’s consider the r2r2r2, if l1−r2l1-r2l1−r2 then l1−r2,l2−r1l1-r2, l2-r1l1−r2,l2−r1 is valid; otherwise li−r2li-r2li−r2 then l1−r1,li−r2l1-r1, li-r2l1−r1,li−r2 is also valid. Therefore, there must has at least 222 matches.
. we choose 222 matches, e.g., l1−r1,l2−r2l1-r1, l2-r2l1−r1,l2−r2, Add a new edge r1→l2r1 \to l2r1→l2; now r1,l2r1, l2r1,l2 is no longer Source or End, so we need remove them from BBB; Operation: 1 remove r1, l2 from BBB 2 Add new edges to BBB (add edge li→rjli \to rjli→rj if li→r1∧l2→rjli \to r1 \land l2 \to rjli→r1∧l2→rj) In fact, the new BBB is the Bipartite of the graph (maybe not DAG) (V,E+(r1→l2)(V, E + (r1 \to l2)(V,E+(r1→l2); the new BBB also has the property: any points has at least one edge; so continue this process until min(∣L∣,∣R∣)<2min( |L|, |R|) < 2min(∣L∣,∣R∣)<2;
2 Now, min(∣L∣,∣R∣)=1min( |L|, |R|) = 1min(∣L∣,∣R∣)=1 (it is impossible that it equals 000, cuz each process above will cause min(∣L∣,∣R∣)−1min( |L|, |R|) - 1min(∣L∣,∣R∣)−1; if ∣L∣=1|L| = 1∣L∣=1, for the property of BBB we have that there have edges l1→ri∀ril1 \to ri \quad \forall ril1→ri∀ri, so we add edge ri→ri+1ri \to r_{i+1}ri→ri+1 and rn→l1rn \to l1rn→l1 (the similar device for the case ∣R∣=1|R| = 1∣R∣=1), then BBB would become a SCC; that is, add max(L,R)max( L, R)max(L,R) edges;
Therefore, the total count is: let kkk be the count of process 1 (add kkk edges, and causes ∣L∣,∣R∣|L|, |R|∣L∣,∣R∣ minus kkk), at 2 we will add max(L−k,R−k)max( L - k, R - k)max(L−k,R−k) edges, so max(L−k,R−k)+k=max(L,R)max( L - k, R - k) + k = max( L, R)max(L−k,R−k)+k=max(L,R)
Pseudocode:
Add_answer( int a, int b); //< add a new edge (a -> b) on G, it would be called `max(n, m)` times.G; //< DAG with Source (s1,...,sn), End (e1,...,em)
B; //< The Bipartite-Graph of `G`
L = {s1, ..., sn}; //< The left-set of `B`
R = {e1, ..., em}; //< The right-set of `B`
for( si){for( ej){if( si can reach to ej){B.add_edge( si, ej);}}
}int l = n, r = m;
while( min( l, r) >= 2){find two matches (denoted `l1-r1, l2-r2`);Add_answer( r1, l2);for( li : L){for( rj : R){if( there is a edge `li -> r1` && `l2 -> rj){B.add_edge( li, rj);}}remove `l1,r2` from `B`;remove `l1` from `L`, remove `r2` from `R`;-- l, -- r;
}// now, min(l, r) = 1
if( l <= r){for( int i = 0; i + 1 < r; ++i){Add_answer( R[i], R[i+1]);}Add_answer( R[r-1], L[0]);
}
else{for( int i = 0; i + 1 < l; ++i){Add_answer( L[i+1], L[i]);}Add_answer( R[0], L[0]);
}
@Delimiter
A Wrong Understanding
A wrong-understanding of the above idea is, let the maximal-match of BBB is kkk, l1−r1,...,lk−rkl1-r1, ..., lk-rkl1−r1,...,lk−rk, so there left (n−k)+(m−k)(n-k) + (m-k)(n−k)+(m−k) unmatched. Therefore, we connect k−1k-1k−1 edges ri→li+1ri \to l_{i+1}ri→li+1, and (n−k)+(m−k)+1(n-k) +(m-k) +1(n−k)+(m−k)+1 edges rk→...→l1rk \to ... \to l1rk→...→l1;
This is wrong, e.g., G=(l1−r1)(l2−r1)(l3−r2)(l3−r3)G = (l1-r1) (l2-r1) (l3-r2) (l3-r3)G=(l1−r1)(l2−r1)(l3−r2)(l3−r3) (its Bipartite-Graph BBB is the same as GGG), the maximal-match is 222, let it be (l1−r1)(l3−r2)(l1-r1) (l3-r2)(l1−r1)(l3−r2), and l2,r3l2, r3l2,r3 are unmatched, then we add edge r1→l3r1 \to l3r1→l3 and r2→l2→r3→l1r2 \to l2 \to r3 \to l1r2→l2→r3→l1 (444 edges in total), surely it also becomes a SCC, but this is not the optimal choice (e.g., r2→l1,r3→l2,r1→l3r2\to l1, r3\to l2, r1 \to l3r2→l1,r3→l2,r1→l3 is also a SCC with 333 edges)
The correct construction-process based on the idea discussed above is,
1 Firstly, we choose two matches (l1−r1)(l3−r2)(l1-r1) (l3-r2)(l1−r1)(l3−r2) of BBB, add a edge r1→l3r1 \to l3r1→l3, subsequently r1,l3r1, l3r1,l3 are no longer the Source/End-Points in the new G1=G+(r1→l3)G1 = G + (r1 \to l3)G1=G+(r1→l3), the new Bipartite-Graph B1B1B1 of the new graph G1G1G1 becomes (l1−r2)(l1−r3)(l2−r2)(l2−r3)(l1-r2) (l1-r3) (l2-r2) (l2-r3)(l1−r2)(l1−r3)(l2−r2)(l2−r3)
2 Secondly, choose two matches (l1−r2)(l2−r3)(l1-r2) (l2-r3)(l1−r2)(l2−r3) of BBB, add a edge r2→l2r2 \to l2r2→l2, the new B2B2B2 of the new graph G2=G1+(r2→l2)G2 = G1 + (r2 \to l2)G2=G1+(r2→l2) becomes l1−r3l1-r3l1−r3
3 Finally, the understanding here is vital. the current BBB is l1−r3l1-r3l1−r3, what does it means? There are only one Source in current GGG, one End; let’s see the current-graph G:(l1)→(SCC)→(r3)G: (l1) \to (SCC) \to (r3)G:(l1)→(SCC)→(r3) where SCC=(l2,l3,r1,r2)SCC = (l2,l3,r1,r2)SCC=(l2,l3,r1,r2), in other words, once we make BBB a SCC, then GGG is also a SCC; Therefore, we add r3→l1r3 \to l1r3→l1 to make BBB a SCC;
One Tips:
At any step, every Closed-Sub-Graphs (Weakly-Connected) of GiGiGi, must has at least oneoneone Sources and Ends;
e.g., a Closed-Sub-Graph in GiGiGi would not be (l1→r1)(r1→l2)(l2→r1)(l1 \to r1) (r1 \to l2) (l2 \to r1)(l1→r1)(r1→l2)(l2→r1) which has 111 Source while 000 End. This case would never happened. Cuz each time we added a edge ri→ljri \to ljri→lj, it means there are li→rili \to rili→ri and lj→rjlj \to rjlj→rj, so in the new Closed-Sub-Graph (after add the new edge) which contains this new edge, it would still contains lilili as its Source and rjrjrj as its End.
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