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;
        }else
            del[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;
}