POJ 1966 Cable TV Network 枚举最小割

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Cable TV Network

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 3740		Accepted: 1753

Description

The interconnection of the relays in a cable TV network is bi-directional. The network is connected if there is at least one interconnection path between each pair of relays present in the network. Otherwise the network is disconnected. An empty network or
a network with a single relay is considered connected. The safety factor f of a network with n relays is: 

1. n, if the net remains connected regardless the number of relays removed from the net. 

2. The minimal number of relays that disconnect the network when removed. 



For example, consider the nets from figure 1, where the circles mark the relays and the solid lines correspond to interconnection cables. The network (a) is connected regardless the number of relays that are removed and, according to rule (1), f=n=3. The network
(b) is disconnected when 0 relays are removed, hence f=0 by rule (2). The network (c) is disconnected when the relays 1 and 2 or 1 and 3 are removed. The safety factor is 2.
Input

Write a program that reads several data sets from the standard input and computes the safety factor for the cable networks encoded by the data sets. Each data set starts with two integers: 0<=n<=50,the number of relays in the net, and m, the number of cables
in the net. Follow m data pairs (u,v), u < v, where u and v are relay identifiers (integers in the range 0..n-1). The pair (u,v) designates the cable that interconnects the relays u and v. The pairs may occur in any order.Except the (u,v) pairs, which do not
contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.
Output

For each data set, the program prints on the standard output, from the beginning of a line, the safety factor of the encoded net.
Sample Input

0 0
1 0
3 3 (0,1) (0,2) (1,2)
2 0
5 7 (0,1) (0,2) (1,3) (1,2) (1,4) (2,3) (3,4)

Sample Output

0
1
3
0
2

Hint

The first data set encodes an empty network, the second data set corresponds to a network with a single relay, and the following three data sets encode the nets shown in figure 1.
Source

Southeastern Europe 2004

给你一个无向图，让你求这个图的安全系数f。
1.如果f为n，不管删除多少点，剩下的点都是连通的。
2.f为删除最少的点使得剩下的图不连通。
求最小点割集。
建图：
将每个点拆成两个，i和i',它们之间的流量为1.
如果两个点相连，则u'->v流量为inf，因为是无向图，所以v'->u流量也是inf。
源点是n'，枚举汇点求最大流。
如果最大流都是inf，那么此图就是连通的，那么f就是n。

//1292K	47MS
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int n,m;
int map[300][300],pre[300],flow[300][300],p[300],a[300];
int EK(int s,int t)
{
int sum=0;
int m=2*s+2;
queue<int>q;
memset(flow,0,sizeof(flow));
for(;;)
{
memset(a,0,sizeof(a));//记录残量
a[s]=inf;
q.push(s);
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=0; i<m; i++)
if(!a[i]&&map[u][i]>flow[u][i])
{
p[i]=u;//记录i的父亲节的是u
q.push(i);
a[i]=a[u]<map[u][i]-flow[u][i]?a[u]:map[u][i]-flow[u][i];
}
}
if(!a[t])break;//如果残量是0的话，就找到最大流
for(int i=t; i!=s; i=p[i])//每条路加上最小残量
{
flow[p[i]][i]+=a[t];
flow[i][p[i]]-=a[t];
}
sum+=a[t];//记录流量
}
return sum;
}
int main()
{
int a,b;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(map,0,sizeof(map));
memset(p,0,sizeof(p));
for(int i=0;i<n;i++)
map[i][i+n]=1;
while(m--)
{
scanf(" (%d,%d)",&a,&b);
map[a+n][b]=inf;
map[b+n][a]=inf;
}
int ans=inf;
for(int i=1;i<n;i++)
ans=min(ans,EK(n,i));
if(ans==inf)ans=n;
printf("%d\n",ans);
}

return 0;
}