poj 1966(无向图点连通度 )

无向图点连通度：

构造有向图，再枚举源汇点求最小割。

1、无向图中任一个点u拆边为u和u+N，连容量为1的有向边。

2、对于无向图中原有的任意无向边(u,v)，在构造的有向图中添加边(u+N,v)和(v+N,v)，容量均为无穷。（拆边规则）

3、O(n^2)枚举源点(拆后出点)、汇点(拆后入点)求最小割的最小值(注意源点与汇点直接相连的情况，显然此时最大流流出来的是inf)。

// File Name: 1966.cpp
// Author: Missa
// Created Time: 2013/4/19 星期五 10:29:17

#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 = 105;
const int maxm = 10010;
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 tmp)
{
    int res = 0;
    while (bfs(st, en))
    {
        for (int i = 0; i <= n; ++i)
            cur[i] = head[i];
        res += dfs(st, inf);
        if (res > tmp) return res;
    }
    return res;
}
int N;
struct Node
{
    int u, v;
}edg[maxm];
void build(int u, int v)
{
    init();
    n = N << 1;
    st = u + N, en = v;
    for (int i = 1; i <= N; ++i)
        addEdge(i, i +N, 1);
    for (int i = 0; i < m; ++i)
    {
        addEdge(edg[i].u + N, edg[i].v, inf);
        addEdge(edg[i].v + N, edg[i].u, inf);
    }
}

int main()
{
    while(~scanf("%d%d",&N,&m))
    {
        for (int i = 0; i < m; ++i)
        {
            scanf(" (%d,%d)", &edg[i].u, &edg[i].v);
            edg[i].u++;
            edg[i].v++;
        }
        int ans = inf;
        for (int i = 1; i <= N; ++i)
            for (int j = 1; j <= N; ++j)
            {
                if ( i == j) continue;
                build(i, j);
                int tmp = dinic(st, en, ans);
                if ( tmp < ans)
                    ans = tmp;
            }
        if (ans == inf) ans = N;
        printf("%d\n",ans);
    }
    return 0;
}