POJ 1815 Friendship 最小割

题目链接：https://vjudge.net/problem/POJ-1815

题意：求s点到t点，最少去掉几个点使得他们不连通。如果无解输出NO ANSWER!

解法：

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn = 1010;
const int maxm = 100010;
const int inf = 0x3f3f3f3f;
int n, st, en;
int start[maxn][maxn];
bool del[maxn];
vector <int> path;
struct G
{int v, cap, next;G() {}G(int v, int cap, int next) : v(v), cap(cap), next(next) {}
} E[maxm];
int p[maxn], T;
int d[maxn], temp_p[maxn], qw[maxn]; //d顶点到源点的距离标号,temp_p当前狐优化，qw队列
void init()
{memset(p, -1, sizeof(p));T = 0;
}
void add(int u, int v, int cap)
{E[T] = G(v, cap, p[u]);p[u] = T++;E[T] = G(u, 0, p[v]);p[v] = T++;
}
bool bfs(int st, int en, int n)
{int i, u, v, head, tail;for(i = 0; i <= n; i++) d[i] = -1;head = tail = 0;d[st] = 0;qw[tail] = st;while(head <= tail){u = qw[head++];for(i = p[u]; i + 1; i = E[i].next){v = E[i].v;if(d[v] == -1 && E[i].cap > 0){d[v] = d[u] + 1;qw[++tail] = v;}}}return (d[en] != -1);
}
int dfs(int u, int en, int f)
{if(u == en || f == 0) return f;int flow = 0, temp;for(; temp_p[u] + 1; temp_p[u] = E[temp_p[u]].next){G& e = E[temp_p[u]];if(d[u] + 1 == d[e.v]){temp = dfs(e.v, en, min(f, e.cap));if(temp > 0){e.cap -= temp;E[temp_p[u] ^ 1].cap += temp;flow += temp;f -= temp;if(f == 0)  break;}}}return flow;
}
void build(){init();for(int i=1; i<=n; i++){if(!del[i]) add(i,i+n,1);}for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)if(start[i][j])add(i+n, j, inf);
}
int dinic(int st, int en, int n)
{int i, ans = 0;while(bfs(st, en, n)){for(i = 0; i <= n; i++) temp_p[i] = p[i];ans += dfs(st, en, inf);}return ans;
}int main(){scanf("%d %d %d", &n,&st,&en);for(int i=1; i<=n; i++)for(int j=1; j<=n; j++)scanf("%d", &start[i][j]);if(start[st][en]){return 0*printf("NO ANSWER!\n");}st = st+n, en = en;build();int ans = dinic(st, en, 2*n+1);printf("%d\n", ans);if(!ans) return 0;for(int i=1; i<=n; i++){//因为割点可能会有很多组合，而且要求字典序最小，所以只能割一个就标记上，一直到割完为止if(i==st-n||i==en) continue;del[i] = 1;build();int t = dinic(st, en, 2*n+1);if(t < ans){ans--;path.push_back(i);if(ans==0) break;}elsedel[i] = false;}for(int i=0; i<path.size(); i++){if(i==path.size()-1) printf("%d\n", path[i]);else printf("%d ", path[i]);}return 0;
}