poj 1815(最小割 割集)

N个人，告诉一系列谁有谁的联系方式。 问题：指定S,T两个人，问除这两个人以外，最少踢掉几个人，可以使得这两个人无法联系的上。给出字典序最小的一种方案。

拆点。因为题目要求的是剔除多少个点。我们要把他转换成剔除多少条边。所以要拆点。

枚举删除每一条边，其余边与原图一致，判断新图的最小割是否比原图减少。减少说明是割集的边。

// File Name: 1815.cpp
// Author: Missa
// Created Time: 2013/4/18 星期四 19:31:30

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<cstdlib>
#include<map>
#include<set>
using namespace std;
#define CL(x,v) memset(x,v,sizeof(x));
#define R(i,st,en) for(int i=st;i<en;++i)
#define LL long long

const int inf = 0x3f3f3f3f;
const int maxn = 410;
const int maxm = 160010;
struct Edge
{
    int v, c, next;//指向点,流量,下一个点
}p[maxm << 1];
int head[maxn], e;
int d[maxn], cur[maxn];//层次记录，当前弧优化
int n, m, st, en;
void init()
{
    e = 0;
    memset(head, -1, sizeof(head));
}
void addEdge(int u, int v, int c)
{
    p[e].v = v; p[e].c = c; 
    p[e].next = head[u];head[u] = e++;
    p[e].v = u; p[e].c = 0; //无向图时逆边赋值流量cap，有向图时赋值0.
    p[e].next = head[v];head[v] = e++;
}
int bfs(int st, int en)
{
    queue <int> q;
    memset(d, 0, sizeof(d));
    d[st] = 1;
    q.push(st);
    while (!q.empty())
    {
        int u = q.front(); q.pop();
        for (int i = head[u]; i != -1; i = p[i].next)
        {
            if (p[i].c > 0 && !d[p[i].v])
            {
                d[p[i].v] = d[u] + 1;
                q.push(p[i].v);
            }
        }
    }
    return d[en];
}
int dfs(int u, int a)//a表示流量
{
    if (u == en || a == 0) return a;
    int f, flow = 0;
    for (int& i = cur[u]; i != -1; i = p[i].next)
    {
        if (d[u] + 1 == d[p[i].v] && (f = dfs(p[i].v,min(a,p[i].c))) > 0)
        {
            p[i].c -= f;
            p[i^1].c += f;
            flow += f;
            a -= f;
            if (!a) break;
        }
    }
    return flow;
}
int dinic(int st, int en)
{
    int res = 0;
    while (bfs(st, en))
    {
        for (int i = 0; i <= n; ++i)
            cur[i] = head[i];
        res += dfs(st, inf);
    }
    return res;
}
int N;
int g[210][210];
bool del[210];
void build()
{
    init();
    for (int i = 1; i <= N; ++i)
    {
        if (del[i]) continue;
        for (int j = 1; j <= N; ++j)
        {
            if (del[j]) continue;
            if (i == j)
                addEdge(i, j + N, 1);
            else if (g[i][j])
                addEdge(i + N, j, inf);
        }
    }
}

int main()
{
    while(~scanf("%d%d%d",&N,&st,&en))
    {
        memset(del, 0, sizeof(del));
        st += N;
        n = N << 1;
        for (int i = 1; i <= N; ++i)
            for(int j = 1; j <= N; ++j)
                scanf("%d", &g[i][j]);
        if(g[st - N][en])
        {
            printf("NO ANSWER!");
            continue;
        }
        build();
        int pre = dinic(st, en);
        int ans = 0;
        for (int i = 1; i <= N; ++i)
        {
            if (pre == 0) break;
            if (i == st - N || i == en) continue;
            del[i] = 1;
            build();
            int tmp = dinic(st, en);
            if (tmp < pre){ans++; pre = tmp;}
            else del[i] = 0;
        }
        printf("%d\n",ans);
        for (int i = 1; i <= N; ++i)
            if(del[i])
                printf("%d ",i);
        printf("\n");
    }
    return 0;
}