学习笔记--网络流24题（上）

题目链接：https://www.luogu.com.cn/problem/P1251

技巧：本题将每个天拆分为两个阶段（白天，晚上）每天晚上会收到脏餐巾（来源：当天早上用完的餐巾，在这道题中可理解为从原点获得），每天早上又有干净的餐巾（来源：购买、快洗店、慢洗店）。

细节：这里在实现的时候，不需要在意买了新的之后才能去洗然后再用于之后，因为即使是先用了洗过之后的也没有关系，流量已经减少了，并且，能用洗过的一定是合法的，三个if已经判断过了。因为要保证最大流，所以一定会去买新的弥补之前不够的

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 2e5 + 5;
int n , st , ed;
int pp , mm , ff , nn , ss;
int head[N], cnt, r[N];
struct node {int f, t, next, w, c;
}edge[N << 1];
void add (int f, int t, int w , int c)
{edge[cnt].f = f;edge[cnt].w = w;edge[cnt].c = c;edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int w, int c)
{add (f, t, w, c);add (t, f, 0, -c);
}
int fw[N], vis[N], pre[N];
ll dis[N];
queue <int> q;
bool spfa()
{mem(dis,INF);mem(fw,INF);mem(vis,0);pre[st] = pre[ed] = -1;q.push(st);vis[st] = 1;dis[st] = 0;while(!q.empty()){int u = q.front();q.pop();vis[u] = 0;for (int i = head[u] ; i != -1 ; i = edge[i].next){int v = edge[i].t;int w = edge[i].w, c = edge[i].c;if(edge[i].w > 0 && dis[u] + c < dis[v]){dis[v] = dis[u] + c;fw[v] = min(fw[u], w);pre[v] = i;if(!vis[v]){q.push(v);vis[v] = 1;}}}}return pre[ed] != -1;
}
ll FYL()
{ll sum = 0 ;while(spfa()){cout << dis[ed] << ' ' << fw[ed] << endl;sum += 1ll * dis[ed] * fw[ed];for (int i = pre[ed] ; i != -1 ; i = pre[edge[i].f]){edge[i].w -= fw[ed];edge[i^1].w += fw[ed];}}return sum;
}
int main ()
{//read (n);mem(head,-1);st = 0, ed = 2 * n + 1;for (int i = 1 ; i <= n ; i ++){read (r[i]);// i 代表早上， i + n 代表晚上addedge (st, i, r[i], 0); //每一天需要的干净餐巾addedge (i + n , ed, r[i], 0); //使用干净的餐巾}read (pp , mm , ff , nn , ss);for (int i = 1 ; i <= n ; i ++){addedge (st, i + n , INF, pp); //直接买干净的餐巾if (i + 1  <= n) addedge (i , i + 1, INF, 0); //不送去洗if (i + mm <= n) addedge (i , i + mm + n , INF , ff); //把早上用的餐巾送到晚上去洗，mm天后可以使用if (i + nn <= n) addedge (i , i + nn + n , INF , ss); //把早上用的餐巾送到晚上去洗，nn天后可以使用}write (FYL()), LF;
}

题目链接：https://www.luogu.com.cn/problem/P2754

技巧：对于一些看起来只有一个点，但是这个点对应着很多状态的时候，一般可以考虑拆点分层图，一般以时间（具体情况具体分析）为轴拆点，分层。这个技巧在建图过程中很常用
实现：枚举时间，对于每一个时间加边建立出符合当前时间的图

图的样子(引用luogu的)：

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int M = 1e6 + 5, N = 105;
int st , ed;
int n , m, k;
int head[M], cnt;
struct node {int t, next, c;
}edge[M << 1];
void add (int f, int t, int c)
{edge[cnt].c = c;edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int c)
{add (f, t, c);add (t, f, 0);
}
int cur[M], level[M];
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{mem(level,0);queue<int>q;q.push(s);level[s] = 1;while(!q.empty()){int root = q.front();q.pop();for(int i = head[root] ; i != -1 ; i = edge[i].next){int v = edge[i].t, c = edge[i].c;if(!level[v] && c > 0){level[v] = level[root] + 1;q.push(v);}}}return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{if(root == ed)return flow;int newflow , cost = 0;for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动{int v = edge[i].t, c = edge[i].c;if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost{cost += newflow;edge[i].c -= newflow;edge[i^1].c += newflow;if(cost == flow) break;}}return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{int ans = 0;while (DinicBfs(st, ed)){for (int i = st ; i <= ed + 5 ; i ++) //如果顶点标号为0，则从0开始cur[i] = head[i];while (int add = DinicDfs(st, INF))ans += add;}return ans;
}
int a[N][N];
int h[N], r[N];
int main ()
{mem (head,-1);read (n, m, k);n += 2;// 地球为2 月亮为 1 其他星球编号集体加2for (int i = 1 ; i <= m ; i ++){read (h[i], r[i]);for (int j = 1 ; j <= r[i] ; j ++){read (a[i][j]);a[i][j] += 2;}}int day = 0, sum = 0;st = 0 , ed = 10005;addedge (st, day * n + 2, INF);// 按时间轴分层，对每一层执行网络流while (day < 500){//对于每个时间的月球都要连接上汇点，源点不用addedge(day * n + 1, ed, INF);if (day){//当day不为0时，每个当前day点往day-1建边，表示停留在该星球for (int i = 1 ; i <= n ; i ++)addedge ((day-1)*n+i, day*n+i, INF);}for (int i = 1; i <= m; i++){//往下一个day状态建边，可以往下继续走，容量为飞船容量int x = a[i][(day + 1) % r[i] + 1];int y = a[i][day % r[i] + 1];addedge((day) * n + y, (day+1) * n + x, h[i]);}//对于每一个时间，跑一次最大流sum += Dinic();if (sum >= k)break;day ++;}if (day == 500)write (0), LF;elsewrite (day) , LF;}

题目链接：https://www.luogu.com.cn/problem/P2756

技巧：比较经典的一对多单向二分图模型，主要是代码理解，fa数组就是对应的链接点

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 2e5 + 5;
int n, m;
int head[N] , cnt;
struct node
{int t, next;
}edge[N];
void add (int f, int t)
{edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
int vis[N] , fa[N];
bool solve (int u)
{for (int i = head[u] ; i != -1 ; i = edge[i].next){int t = edge[i].t;if (!vis[t]){vis[t] = 1;if (!fa[t] || solve (fa[t])){fa[t] = u;return true;}}}return false;
}
int main ()
{mem (head,-1);read (m, n);while (1){int u , v;read (u, v);if (u == -1 && v == -1)break;add (u, v);}int ans = 0;for (int i = 1 ; i <= m ; i ++){mem(vis, 0);if (solve (i))ans ++;}cout << ans << endl;for (int i = 1 ; i <= n ; i ++)if (fa[i])cout << fa[i] << ' ' << i << endl;
}

题目链接：https://www.luogu.com.cn/problem/P2761

技巧：比较典型的状态压缩+最短路，读懂题目之后直接模拟即可

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int M = (1 << 21) - 1, N = 205;
int n , m;
struct node {int b1 = 0, b2 = 0;int f1 = 0, f2 = 0;int w;
}p[N];
int vis[M] , dis[M];
void spfa ()
{queue <int> q;int s = (1 << n) - 1;mem(dis, INF);dis[s] = 0;vis[s] = 1;q.push(s);while (!q.empty()){int u = q.front();q.pop ();vis[u] = 0;for (int i = 1 ; i <= m ; i ++){if (((u&p[i].b1)==p[i].b1)&&((u&p[i].b2)==0)){int v = (((u | p[i].f1) | p[i].f2) ^ p[i].f1);if (p[i].w + dis[u] < dis[v]){dis[v] = p[i].w + dis[u];if (!vis[v]){q.push(v);vis[v] = 1;}}}}}
}
int main ()
{read (n, m);for (int i = 1 ; i <= m ; i ++){read (p[i].w);string s1 , s2;cin >> s1 >> s2;for (int j = 0 ; j < s1.length() ; j ++) {if (s1[j] == '+')p[i].b1 += 1 << j;else if (s1[j] == '-')p[i].b2 += 1 << j;}for (int j = 0 ; j < s2.length() ; j ++) {if (s2[j] == '-')p[i].f1 += 1 << j;else if (s2[j] == '+')p[i].f2 += 1 << j;}}spfa ();if (dis[0] == INF)cout << "0" << endl;elsecout << dis[0] << endl;
}

题目链接：https://www.luogu.com.cn/problem/P2762

技巧：最大权闭合子图经典题目，由于题目描述：如果选择了某个实验项目，那么实验器材也必须选，很明显的闭合子图性质。
细节：这里一定要注意，在构造闭合子图的时候，不可以只通过判断与源点相链的边得容量。需要从源点出发搜索，可达点均为闭合子图中的点

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 1e6 + 5;
int n , m , head[N] , cnt;
struct node {int t, next, c;
}edge[N << 1];
void add (int f, int t, int c)
{edge[cnt].t = t;edge[cnt].c = c;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int c)
{add (f, t, c);add (t, f, 0);
}
int st , ed;
int cur[N], level[N], vis[N];
vector <int> a[N];
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{mem(level,0);queue<int>q;q.push(s);level[s] = 1;while(!q.empty()){int root = q.front();q.pop();for(int i = head[root] ; i != -1 ; i = edge[i].next){int v = edge[i].t, c = edge[i].c;if(!level[v] && c > 0){level[v] = level[root] + 1;q.push(v);}}}return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{if(root == ed)return flow;int newflow , cost = 0;for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动{int v = edge[i].t, c = edge[i].c;if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost{cost += newflow;edge[i].c -= newflow;edge[i^1].c += newflow;if(cost == flow) break;}}return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{int ans = 0;while (DinicBfs(st, ed)){for (int i = st ; i <= ed + 5 ; i ++) //如果顶点标号为0，则从0开始cur[i] = head[i];while (int add = DinicDfs(st, INF))ans += add;}return ans;
}
char ch[N];
int ulen, num;
int main ()
{mem (head, -1);int sum = 0;read (m, n);st = 0 , ed = n + m + 1;for (int i = 1 ; i <= m ; i ++){int p;read (p);sum += p;addedge (st, i , p);memset(ch,0,sizeof(ch));cin.getline(ch,10000);ulen=0;while (sscanf(ch+ulen,"%d",&num)==1){a[i].pb(num);addedge (i , m + num , INF);if (num==0) ulen++;else while (num){num/=10;ulen++;}ulen++;}}for (int i = 1 ; i <= n ; i ++){int p;read (p);addedge (m + i , ed , p);}int ans = Dinic();//这里找出残余网络中S可到的点//这里不能直接检查与S相连的边的容量，因为可能经过实验器材之后再走到实验for (int i = 1 ; i <= m ; i ++){if (level[i] != 0) {cout << i << ' ';vis[i] = 1;}}cout << endl;set <int> s;for (int i = 1 ; i <= m ; i ++){if (vis[i]) {for (auto v : a[i])s.insert (v);}}for (auto v : s)cout << v << ' ';cout << endl;cout << sum - ans << endl;}

题目链接：https://www.luogu.com.cn/problem/P2763

技巧：无
细节：无

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 1e6 + 5;
int k , n , st , ed , m;
int a[N];
int head[N], cnt;
struct node {int t, next, c;
}edge[N << 1];
void add (int f, int t, int c)
{edge[cnt].c = c;edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int c)
{add (f, t, c);add (t, f, 0);
}
int cur[N], level[N];
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{mem(level,0);queue<int>q;q.push(s);level[s] = 1;while(!q.empty()){int root = q.front();q.pop();for(int i = head[root] ; i != -1 ; i = edge[i].next){int v = edge[i].t, c = edge[i].c;if(!level[v] && c > 0){level[v] = level[root] + 1;q.push(v);}}}return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{if(root == ed)return flow;int newflow , cost = 0;for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动{int v = edge[i].t, c = edge[i].c;if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost{cost += newflow;edge[i].c -= newflow;edge[i^1].c += newflow;if(cost == flow) break;}}return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{int ans = 0;while (DinicBfs(st, ed)){for (int i = st ; i <= ed + 5 ; i ++) //如果顶点标号为0，则从0开始cur[i] = head[i];while (int add = DinicDfs(st, INF))ans += add;}return ans;
}
int main ()
{mem (head, -1);read (k , n);st = 0 , ed = k * n + k + 1;for (int i = 1 ; i <= k ; i ++){read(a[i]);m += a[i];addedge (k * n + i , ed, a[i]);}for (int i = 1 ; i <= n ; i ++){cout << i << ":" << endl;int p;read (p);for (int j = 0 ; j < p ; j ++) {int x;read (x);cout << i + (x-1) * n << ' ' << k * n + x << endl;addedge (st , i + (x-1) * n , 1);addedge (i + (x-1) * n, k * n + x , 1);}cout << endl;}int ans = Dinic ();if (ans != m)puts ("No Solution!");else{for (int u = k * n + 1 ; u <= k * n + k ; u ++){printf ("%d:", u % n);for (int i = head[u] ; i != -1 ; i = edge[i].next){int v = edge[i].t , c = edge[i].c;if (v != ed && c == 1) {if (v % n == 0)printf (" %d",n);elseprintf(" %d",v%n);}}printf ("\n");}}
}

题目链接：https://www.luogu.com.cn/problem/P2764
技巧：对原图中的每一个点拆点为(x , x + n)，如果存在一条边(x, y)则，新图中存在 (x , y + n)，那么 最少路径数 = 原图点数 - 新图最大匹配数，同时，每一条路径的终点等于新图中未匹配到的点.,该类问题属于 最少不相交路径覆盖问题

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}
const int N = 20005;
int head[N], cnt;
int n , m;
struct node {int t, next;
}edge[N];
void add (int f, int t)
{edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
int vis[N] , fa[N], t[N];
bool solve (int u)
{for (int i = head[u] ; i != -1 ; i = edge[i].next){int t = edge[i].t;if (!vis[t]){vis[t] = 1;if (!fa[t] || solve (fa[t])){fa[t] = u;return true;}}}return false;
}
vector <int> Edge[N];
vector <int> path;
void dfs (int u)
{path.pb(u);for (auto v : Edge[u]){if (!vis[v]){vis[v] = 1;dfs (v);}}
}
int main ()
{read (n , m);mem (head, -1);for (int i = 1 ; i <= m ; i++){int u , v;read (u, v);Edge[v].pb(u);add (u , v + n + 1);}int ans = 0;for (int i = 1 ; i <= n ; i ++){mem(vis, 0);if (solve (i)) {t[i] = 1;ans++;}}for (int i = 1 ; i <= n ; i ++){if (!t[i]) {vis[i] = 1;//cout << i << endl;path.clear();dfs(i);reverse (path.begin(), path.end());for (auto v : path)cout << v << ' ';cout << endl;}}cout << n - ans << endl;
}

题目链接：https://www.luogu.com.cn/problem/P2765
技巧：隐式图，图的特征不是很明显，但是可以分析出来，设有x个点，俩俩能组成平方数的数字连边，从而建立出图（按照求最少路径覆盖的方式拆点建图），接下来就是最少路径覆盖了

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int M = 3e6 + 5;
const int Base = 1e6+5;
int n, a[2005], mark[M];
int head[M] , cnt;
struct node {int t, next;
}edge[M << 1];
void add (int f, int t)
{edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
int vis[M] , fa[M], t[M], m , ans;
bool solve (int u)
{for (int i = head[u] ; i != -1 ; i = edge[i].next){int t = edge[i].t;if (!vis[t]){vis[t] = 1;if (!fa[t] || solve (fa[t])){fa[t] = u;return true;}}}return false;
}
vector <int> path, Edge[M];
void dfs (int u)
{path.pb(u);for (auto v : Edge[u]){if (v == m + 1)continue;if (!vis[v]){vis[v] = 1;dfs (v);break;}}
}
int main ()
{mem (head, -1);int dd = 1;for (int i = 1 ; i <= 1000 ; i ++){mark[dd * dd] = 1;dd += 1;}read (n);m = 0, ans = 0;while (1){m ++;for (int i = 1 ; i < m ; i ++)if (mark[i + m]) {add(m, i + Base);Edge[i].pb(m);}mem (vis, 0);int p = solve (m);ans += p;if (p)t[m] = 1;if (m - ans > n)break;}mem (vis, 0);cout << --m << endl;for (int i = 1 ; i <= m ; i ++){if (!t[i]){vis[i] = 1;path.clear();dfs(i);for (auto v : path)cout << v << ' ';cout << endl;}}
}

题目链接：https://www.luogu.com.cn/problem/P2766
技巧：拆点和建图的思想很精妙，通过dp数组和原数组建立出关系图，然后利用拆点作为限制，流量的控制满足不同题意

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 10005;
int n, a[N], dp[N];
int ans, head[N], cnt, st , ed;
struct node {int next, t, c;
}edge[N << 3];
void add (int f, int t, int c)
{edge[cnt].t = t;edge[cnt].c = c;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int c)
{add (f , t, c);add (t , f, 0);
}
int cur[N], level[N];
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{mem(level,0);queue<int>q;q.push(s);level[s] = 1;while(!q.empty()){int root = q.front();q.pop();for(int i = head[root] ; i != -1 ; i = edge[i].next){int v = edge[i].t, c = edge[i].c;if(!level[v] && c > 0){level[v] = level[root] + 1;q.push(v);}}}return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{if(root == ed)return flow;int newflow , cost = 0;for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动{int v = edge[i].t, c = edge[i].c;if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost{cost += newflow;edge[i].c -= newflow;edge[i^1].c += newflow;if(cost == flow) break;}}return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{int ans = 0;while (DinicBfs(st, ed)){for (int i = st ; i <= ed + 5 ; i ++) //如果顶点标号为0，则从0开始cur[i] = head[i];while (int add = DinicDfs(st, INF))ans += add;}return ans;
}
int main ()
{mem (head,-1);read (n);if (n == 1){cout << 1 << endl;cout << 1 << endl;cout << 1 << endl;return 0;}for (int i = 1 ; i <= n ; i ++)read (a[i]);for (int i = 1 ; i <= n ; i ++){for (int j = 1 ; j < i ; j ++)if (a[j] <= a[i] && dp[j] > dp[i])dp[i] = dp[j];dp[i] ++;ans = max (dp[i], ans);}st = 0 , ed = 2 * n + 1;for(int i = 1 ; i <= n ; i ++){addedge (i , i + n , 1);if (dp[i] == 1)addedge(st, i, 1);// 不能是 else if 因为可能 1 == ansif (dp[i] == ans)addedge(i + n, ed, 1);}for(int i = 1 ; i <= n ; i ++)for(int j = 1 ; j < i ; j ++)if(a[j] <= a[i] && dp[j] == dp[i]-1)addedge(j + n , i ,1);int tot = Dinic ();cout << ans << endl;cout << tot << endl;addedge(1,1+n,INF), addedge(st,1,INF);if(dp[n]==ans) addedge(n+n,ed,INF),addedge(n,n+n,INF);tot += Dinic ();cout << tot << endl;}

题目链接：https://www.luogu.com.cn/problem/P2770

技巧：最大费用最大流，拆点限制行走的次数，构造答案的时候对残余网络进行搜索即可

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)
#define sz(a) (int)a.size()template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 100010;
int n , m, maxflow;
string str[N];
map <string, int> ma;
int head[N], cnt;
int st , ed;
struct node {int f, t, next, w, c;
}edge[N * 20];
void add (int f, int t, int w , int c)
{edge[cnt].f = f;edge[cnt].w = w;edge[cnt].c = c;edge[cnt].t = t;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int w, int c)
{add (f, t, w, c);add (t, f, 0, -c);
}
int fw[N], vis[N], pre[N];
ll dis[N];
queue <int> q;
bool spfa()
{for (int i = 0; i <= ed + 5 ; i ++)dis[i] = -INF;mem(fw,INF);mem(vis,0);pre[st] = pre[ed] = -1;q.push(st);vis[st] = 1;dis[st] = 0;while(!q.empty()){int u = q.front();q.pop();vis[u] = 0;for (int i = head[u] ; i != -1 ; i = edge[i].next){int v = edge[i].t;int w = edge[i].w, c = edge[i].c;if(edge[i].w  && dis[u] + c > dis[v]){dis[v] = dis[u] + c;fw[v] = min(fw[u], w);pre[v] = i;if(!vis[v]){q.push(v);vis[v] = 1;}}}}return pre[ed] != -1;
}
ll FYL()
{ll sum = 0 ;while(spfa()){maxflow ++;sum += 1ll * dis[ed] * fw[ed];for (int i = pre[ed] ; i != -1 ; i = pre[edge[i].f]){edge[i].w -= fw[ed];edge[i^1].w += fw[ed];}}return sum;
}
void dfs1 (int u)
{vis[u] = 1;cout << str[u - n] << endl;for (int i = head[u] ; i != -1 ; i = edge[i].next){int t = edge[i].t, w = edge[i].w;if (t <= n && w == 0) {dfs1(t + n);break; //break位置要注意}}
}
void dfs2 (int u)
{vis[u] = 1;for (int i = head[u] ; i != -1 ; i = edge[i].next){int t = edge[i].t, w = edge[i].w;if (t <= n && w == 0 && !vis[t + n])dfs2 (t + n);}cout << str[u - n] << endl;
}
int main ()
{int check = 0;mem (head, -1);cin >> n >> m;for (int i = 1 ; i <= n ; i ++){cin >> str[i];ma[str[i]] = i;}for (int i = 2 ; i <= n - 1 ; i ++)addedge (i , i + n , 1, 1);addedge (1 , 1 + n , 2, 1);addedge (n , n + n , 2, 1);for (int i = 1 ; i <= m ; i ++){string u , v;cin >> u >> v;int x = ma[u] , y = ma[v];if (x > y)swap (x, y);if (x == 1 && y == n)check = 1;addedge (x + n, y, 1, 0);}st = 1 , ed = 2 * n;ll cost = FYL();mem (vis, 0);if (maxflow == 2){cout << cost - 2 << endl;dfs1 (st + n);dfs2 (st + n);}else if (maxflow == 1 && check){cout << 2 << endl;cout << str[1] << endl;cout << str[n] << endl;cout << str[1] << endl;}elsecout << "No Solution!" << endl;
}

题目链接：https://www.luogu.com.cn/problem/P2774

技巧：二分图建图，求最大权独立集合

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)
#define sz(a) (int)a.size()template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}
const int N = 1005;
int n , m, sum = 0;
int num[N][N];
int head[N * N] , cnt;
struct node {int c, t, next;
}edge[N * N * 10];
void add (int f, int t, int c)
{edge[cnt].t = t;edge[cnt].c = c;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int c)
{add (f, t, c);add (t, f, 0);
}
int st, ed, cur[N], level[N];
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{mem(level,0);queue<int>q;q.push(s);level[s] = 1;while(!q.empty()){int root = q.front();q.pop();for(int i = head[root] ; i != -1 ; i = edge[i].next){int v = edge[i].t, c = edge[i].c;if(!level[v] && c > 0){level[v] = level[root] + 1;q.push(v);}}}return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{if(root == ed)return flow;int newflow , cost = 0;for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动{int v = edge[i].t, c = edge[i].c;if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost{cost += newflow;edge[i].c -= newflow;edge[i^1].c += newflow;if(cost == flow) break;}}return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{int ans = 0;while (DinicBfs(st, ed)){for (int i = st ; i <= ed + 5 ; i ++) //如果顶点标号为0，则从0开始cur[i] = head[i];while (int add = DinicDfs(st, INF))ans += add;}return ans;
}
int main ()
{mem (head, -1);read (n, m);st = 0 , ed = n * m + 1;for (int i = 1 ; i <= n ; i ++){for (int j = 1 ; j <= m ; j ++){read (num[i][j]);sum += num[i][j];int id = (i-1) * m + j;//cout << id << endl;if ((i + j) % 2){addedge(st, id, num[i][j]);//下面的建边，一定只能单向，不然的话，可能最小割直接割与ed连接的边就好了//但是由于双向之后，可能只割与ed连的边都不够if (i > 1)addedge (id , id - m , INF);if (i < n)addedge (id , id + m , INF);if (j > 1)addedge (id , id - 1 , INF);if (j < m)addedge (id , id + 1 , INF);}elseaddedge (id , ed, num[i][j]);}}cout << sum - Dinic() << endl;
}

题目链接：https://www.luogu.com.cn/problem/P3254

技巧：典型的二分图模型，求最大匹配，构造输出即可

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)
#define sz(a) (int)a.size()template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);
}ll lcm(ll a, ll b) {return a / gcd(a, b) * b;
}const int N = 50005;
int n , m, r[N], c[N];
int head[N] , cnt, st, ed , cur[N], level[N];
struct node {int t, next , c;
}edge[N * 20];
void add (int f, int t, int C)
{edge[cnt].t = t;edge[cnt].c = C;edge[cnt].next = head[f];head[f] = cnt ++;
}
void addedge (int f, int t, int C)
{add (f, t, C);add (t, f, 0);
}
bool DinicBfs(int s, int e) //利用广搜给网络图分层
{mem(level,0);queue<int>q;q.push(s);level[s] = 1;while(!q.empty()){int root = q.front();q.pop();for(int i = head[root] ; i != -1 ; i = edge[i].next){int v = edge[i].t, c = edge[i].c;if(!level[v] && c > 0){level[v] = level[root] + 1;q.push(v);}}}return level[e] != 0; //如果当前网络能到达汇点说明存在增广路
}
int DinicDfs(int root , int flow) //利用DFS在已经分好层的网络中寻找增广路
{if(root == ed)return flow;int newflow , cost = 0;for(int& i = cur[root] ; i != -1 ; i = edge[i].next) //当前弧优化，&i代表i改动cur也会改动{int v = edge[i].t, c = edge[i].c;if(level[v] == level[root] + 1 && c > 0 && (newflow = DinicDfs(v,min(c,flow-cost))))//多路增广flow-cost{cost += newflow;edge[i].c -= newflow;edge[i^1].c += newflow;if(cost == flow) break;}}return cost; //返回当前分层网络中能够寻找到的最大流
}
int Dinic()
{int ans = 0;while (DinicBfs(st, ed)){for (int i = st ; i <= ed + 5 ; i ++) //如果顶点标号为0，则从0开始cur[i] = head[i];while (int add = DinicDfs(st, INF))ans += add;}return ans;
}
vector <int> path[N];
int main ()
{mem (head, -1);int sum = 0;read (n, m);st = 0 , ed = n + m + 1;for (int i = 1 ; i <= n ; i ++){read (r[i]);sum += r[i];addedge (st, i , r[i]);}for (int i = 1 ; i <= m ; i  ++){read (c[i]);addedge (n + i , ed, c[i]);}for (int i = 1 ; i <= n ; i ++)for (int j = 1 ; j <= m ; j ++)addedge (i , n + j , 1);int ans = Dinic();if (ans == sum){write (1), LF;for (int i = n + 1 ; i <= n + m ; i ++){for (int j = head[i] ; j != -1 ; j = edge[j].next){int t = edge[j].t, w = edge[j].c;if (t != ed && w == 1)path[t].pb(i);}}for (int i = 1 ; i <= n ; i ++){for (auto v : path[i])write (v-n), SP;LF;}}elsewrite (0) , LF;return 0;
}

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