图论模板——最大流问题

Dinic算法的时间复杂度为O（n^2m）。实际运用远远小于这个上界。
特别的，Dinic算法求解二分图最大匹配的时间复杂度为O（msqrt（n））
最大流问题模板

#include <bits/stdc++.h>
using namespace std;
const int N = 10010,M = 200010, INF = 1e8;
int n,m,S,T;
int h[N],ne[M],e[M],f[M],idx;
int d[N],q[N],cur[N];
void add(int a,int b,int c)
{e[idx] = b,f[idx] = c,ne[idx] = h[a],h[a] = idx++;e[idx] = a,f[idx] = 0,ne[idx] = h[b],h[b] = idx++;
}
bool bfs()
{int hh = 0, tt = 0;memset(d, -1, sizeof d);q[0] = S, d[S] = 0, cur[S] = h[S];while (hh <= tt){int t = q[hh ++ ];for (int i = h[t]; ~i; i = ne[i]){int ver = e[i];if (d[ver] == -1 && f[i]){d[ver] = d[t] + 1;cur[ver] = h[ver];if (ver == T)  return true;q[ ++ tt] = ver;}}}return false;
}
int find(int u,int limit)
{if (u==T) return limit;int flow = 0;for (int i = cur[u];~i&&flow<limit;i = ne[i]){cur[u] = i;//当前弧优化int ver = e[i];if (d[ver] == d[u]+1&&f[i]){int t = find(ver,min(f[i],limit-flow));if (!t) d[ver] = -1;f[i] -= t,f[i^1] += t,flow+=t;}}return flow;
}
int dinic()
{int r = 0,flow;while (bfs()) while (flow = find(S,INF)) r += flow;return r;
}
int main()
{scanf("%d%d%d%d",&n,&m,&S,&T);memset(h,-1,sizeof h);while (m--){int a,b,c;scanf("%d%d%d",&a,&b,&c);add(a,b,c);}printf("%d\n",dinic());return 0;
}

应用一：二分图匹配问题
飞行员配对方案问题 https://www.acwing.com/problem/content/2177/

int main()
{scanf("%d%d",&m,&n);S = 0,T = n+1;memset(h,-1,sizeof h);for (int i = 1;i<=m;i++) add(S,i,1);for (int i = m+1;i<=n;i++) add(i,T,1);int a,b;while (scanf("%d%d",&a,&b),a!=-1) add(a,b,1);printf("%d\n",dinic());for (int i=0;i<idx;i+=2)if (e[i]>m&&e[i]<=n&&!f[i]) printf("%d %d\n",e[i^1],e[i]);return 0;
}

应用二：无源汇的最大流问题
无源汇上下界可行流 https://www.acwing.com/problem/content/2190/)
注意流量守恒

int main()
{memset(h,-1,sizeof h);scanf("%d%d",&n,&m);S = 0,T = n+1,tot = 0;for (int i=0;i<m;i++){int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);add(a,b,c,d);A[a] -= c,A[b] += c;//出去c，进来c}for (int i=1;i<=n;i++){if (A[i]>0) add(S,i,0,A[i]),tot += A[i];//如果进来大于0else if (A[i]<0) add(i,T,0,-A[i]);//如果出去}if (dinic()!=tot) printf("NO\n");//如果最低的流量都不行else {printf("YES\n");for (int i=0;i<m*2;i+=2)printf("%d\n",f[i^1]+l[i]);//输出方案}return 0;
}

应用三：有源汇的最大流问题
有源汇上下界最大流 https://www.acwing.com/problem/content/2191/
先链接一个t到s的边，看是否能满足所有边的最小下限
如果满足再做一次dinic()

int main()
{memset(h,-1,sizeof h);int s,t,tot = 0;scanf("%d%d%d%d",&n,&m,&s,&t);S = 0,T = n+1;while (m--){int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);add(a,b,d-c);A[a] -= c,A[b] += c;}for (int i=1;i<=n;i++)if (A[i]>0) add(S,i,A[i]),tot += A[i];else if (A[i]<0) add(i,T,-A[i]);add(t,s,INF);if (dinic()<tot) puts("No Solution");else {int res = f[idx-1];f[idx-1] = f[idx-2] = 0;S = s,T = t;printf("%d",res+dinic());}
}

应用四：有源汇的最小流问题
有源汇上下界最小流 https://www.acwing.com/problem/content/2192/
加t到s的边，然后dinic，然后再反向dinic

int main()
{memset(h,-1,sizeof h);int s,t,tot = 0;scanf("%d%d%d%d",&n,&m,&s,&t);S = 0,T = n+1;while (m--){int a,b,c,d;scanf("%d%d%d%d",&a,&b,&c,&d);add(a,b,d-c);A[a] -= c,A[b] += c;}for (int i=1;i<=n;i++)if (A[i]>0) add(S,i,A[i]),tot += A[i];else if (A[i]<0) add(i,T,-A[i]);add(t,s,INF);if (dinic()<tot) puts("No Solution");else {int res = f[idx-1];f[idx-1] = f[idx-2] = 0;S = t,T = s;printf("%d",res-dinic());}
}

应用五：多源汇的最大流问题
多源汇最大流 https://www.acwing.com/problem/content/2236/

int main()
{memset(h,-1,sizeof h);int ss,tt;scanf("%d%d%d%d",&n,&m,&ss,&tt);S = 0,T = n+1;while (ss--){int x;scanf("%d",&x);add(S,x,INF);}while (tt--){int x;scanf("%d",&x);add(x,T,INF);}while (m--){int a,b,c;scanf("%d%d%d",&a,&b,&c);add(a,b,c);}printf("%d\n",dinic());
}

应用六：关键边
伊基的故事 I - 道路重建 https://www.acwing.com/problem/content/2238/
dinic后从S和T遍历

void dfs(int u, bool st[], int t)
{st[u] = true;for (int i = h[u]; ~i; i = ne[i]){int j = i ^ t, ver = e[i];if (f[j] && !st[ver])dfs(ver, st, t);}
}
int main()
{memset(h,-1,sizeof h);scanf("%d%d",&n,&m);for (int i = 0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);add(a,b,c);}S = 0,T = n-1;dinic();dfs(S, vis_s, 0);dfs(T, vis_t, 1);int res = 0;for (int i = 0; i < m * 2; i += 2)if (!f[i] && vis_s[e[i ^ 1]] && vis_t[e[i]])res ++ ;printf("%d\n", res);return 0;
}

应用七：有边大小使用限制的最大流判定
秘密挤奶机 https://www.acwing.com/problem/content/2279/

bool check(int x)
{for (int i=0;i<idx;i++) if (w[i]>x) f[i] = 0;else f[i] = 1;return dinic()>=k;
}
int main()
{memset(h,-1,sizeof h);scanf("%d%d%d",&n,&m,&k);S = 1,T = n;for (int i = 0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);add(a,b,c);}int l = 1,r = 1e6;while (l<r){int mid = l+r>>1;if (check(mid)) r = mid;else l = mid+1;}printf("%d\n",r);return 0;
}

应用八：拆点
餐饮 https://www.acwing.com/problem/content/2242/

int main()
{memset(h,-1,sizeof h);scanf("%d%d%d",&n,&m,&k);S = 0,T = 2*n+m+k+1;for (int i = 2*n+1;i<=2*n+m;i++) add(S,i,1);for (int i = 1;i<=n;i++) add(i,i+n,1);for (int i = 2*n+m+1;i<=2*n+m+k;i++) add(i,T,1);for (int i = 1;i<=n;i++){int a,b,x;scanf("%d%d",&a,&b);for (int j = 0;j<a;j++){scanf("%d",&x);add(2*n+x,i,1);}for (int j = 0;j<b;j++){scanf("%d",&x);add(n+i,2*n+m+x,1);}}printf("%d\n",dinic());return 0;
}