题目描述：给你N个节点，和节点之间的关联关系。然后给你其中两个不同的节点S，T。问你最少去掉多少个节点可以使S，T之间没有关联。输出这些点，如果有多个解，输出一个字典序最小的。

解题报告：

拆点，点i分为i和i’，连接i到i’，容量为1。（如果是S和T点，容量为无穷）

如果点a到点b有边，连接a’ 到b，这样求最小割就是最少的节点数。

然后就是怎样寻找答案所需的最优的割点了。

枚举割点：

按字典序枚举到点i，就重新构图（不要点i和之前已经删除的点），如果这样的图求的的最大流小于原来的值，就说明这是一个有效割点，保存之。

这样就是答案。

代码如下：

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

#define Max 0x1fffffff

#define size 500

struct edge{int from, to, val, next;}e[140000];

int v[size], que[size], dis[size], cnt, cur[size];

void insert(int from, int to, int va)

{

    e[cnt].from = from, e[cnt].to = to; e[cnt].val = va;

    e[cnt].next = v[from];v[from] = cnt++;

    e[cnt].from = to, e[cnt].to = from; e[cnt].val = 0;

    e[cnt].next = v[to];v[to] = cnt++;

}

bool bfs(int n, int s, int t)

{

    int head, tail, id;

    head = tail = 0; que[tail++] = s;

    memset(dis, -1, sizeof(int) * n);dis[s] = 0;

    while(head < tail) // bfs,得到顶点i的距s的最短距离dis[i]

        for(id = v[que[head++]]; id != -1; id = e[id].next)

            if (e[id].val > 0 && dis[e[id].to] == -1)

            {

                dis[e[id].to] = dis[e[id].from] + 1;

                que[tail++] = e[id].to;

                if (e[id].to == t) return true;

            }

    return false;

}

int Dinic(int n, int s, int t)

{

    int maxflow = 0, tmp, i;

    while(bfs(n, s, t))

    {

        int u = s, tail = 0;

        for(i = 0; i < n; i++) cur[i] = v[i];

        while(cur[s] != -1)

            if (u != t && cur[u] != -1 && e[cur[u]].val > 0 && dis[u] != -1 && dis[u] + 1 == dis[e[cur[u]].to])

            {que[tail++] = cur[u]; u = e[cur[u]].to;}

            else if (u == t)

            {

                for(tmp = Max, i = tail - 1; i >= 0; i--) tmp = min(tmp, e[que[i]].val);

                for(maxflow += tmp, i = tail - 1; i >= 0; i--)

                {

                    e[que[i]].val -= tmp;

                    e[que[i] ^ 1].val += tmp;

                    if (e[que[i]].val == 0) tail = i;

                }

                u = e[que[tail]].from;

            }

            else

            {

                while(tail > 0 && u != s && cur[u] == -1) u = e[que[--tail]].from;

                cur[u] = e[cur[u]].next;

            }

    }

    return maxflow;

}

int n, s, t, num, jeo[size], ans[size], x[size][size];

bool vst[size];

int main()

{

    while(scanf("%d%d%d", &n, &s, &t) != EOF)

    {

        memset(v, -1, sizeof(int) * (n + n + 1)); cnt = 0;

        for(int i = 1; i <= n; i++)

            if (i != s && i != t) insert(i, i + n, 1);

            else insert(i, i + n, Max);

        for(int i = 1; i <= n; i++)

            for(int j = 1; j <= n; j++)

            {

                scanf("%d", &x[i][j]);

                if (i == j || j == s || i == t) continue;

                if (x[i][j]) insert(i + n, j, Max);

            }

        int jeogia = Dinic(n + n + 1, s, t + n); num = 0;

        if (jeogia == Max) {printf("NO ANSWER!\n"); continue;}

        memset(vst, 0, sizeof(bool) * (n + 1));

        for(int i = 1; i <= n; i++)

        {

            if (i == t || i == s) continue;

            vst[i] = 1;

            memset(v, -1, sizeof(int) * (n + n + 1)); cnt = 0;

            for(int j = 1; j <= n; j++)

                if (vst[j]) continue;

                else if (j != s && j != t) insert(j, j + n, 1);

                else insert(j, j + n, Max);

            for(int j = 1; j <= n; j++)

                for(int k = 1; k <= n; k++)

                {

                    if (k == j || k == s || j == t || vst[j] || vst[k]) continue;

                    if (x[j][k]) insert(j + n, k, Max);

                }

            int tmp = Dinic(n + n + 1, s, t + n);

            if (tmp < jeogia)

            {

                jeo[num++] = i;

                jeogia = tmp;

            }

            else vst[i] = 0;

        }

        printf("%d\n", num);

        for(int i = 0; i < num; i++)

        {

            printf("%d", jeo[i]);

            if (i == num - 1) printf("\n");

            else printf(" ");

        }

    }

    return 0;

}