poj 1815(最大流来求割点)

Friendship

Time Limit: 2000MS		Memory Limit: 20000K
Total Submissions: 5906		Accepted: 1622

Description

In modern society, each person has his own friends. Since all the people are very busy, they communicate with each other only by phone. You can assume that people A can keep in touch with people B, only if
1. A knows B's phone number, or
2. A knows people C's phone number and C can keep in touch with B.
It's assured that if people A knows people B's number, B will also know A's number.

Sometimes, someone may meet something bad which makes him lose touch with all the others. For example, he may lose his phone number book and change his phone number at the same time.

In this problem, you will know the relations between every two among N people. To make it easy, we number these N people by 1,2,...,N. Given two special people with the number S and T, when some people meet bad things, S may lose touch with T. Your job is to compute the minimal number of people that can make this situation happen. It is supposed that bad thing will never happen on S or T.

Input

The first line of the input contains three integers N (2<=N<=200), S and T ( 1 <= S, T <= N , and S is not equal to T).Each of the following N lines contains N integers. If i knows j's number, then the j-th number in the (i+1)-th line will be 1, otherwise the number will be 0.

You can assume that the number of 1s will not exceed 5000 in the input.

Output

If there is no way to make A lose touch with B, print "NO ANSWER!" in a single line. Otherwise, the first line contains a single number t, which is the minimal number you have got, and if t is not zero, the second line is needed, which contains t integers in ascending order that indicate the number of people who meet bad things. The integers are separated by a single space.

If there is more than one solution, we give every solution a score, and output the solution with the minimal score. We can compute the score of a solution in the following way: assume a solution is A1, A2, ..., At (1 <= A1 < A2 <...< At <=N ), the score will be (A1-1)*N^t+(A2-1)*N^(t-1)+...+(At-1)*N. The input will assure that there won't be two solutions with the minimal score.

Sample Input

3 1 3
1 1 0
1 1 1
0 1 1

Sample Output

1
2

  如果单纯是求割点数的话。。。这个题不难。。直接拆点然后最大流处理一下就可以了。。问题是这个题要求按照字典序（依题意推出来的）把最少的割点输出来。。。。。大概思路就是 。。。枚举每个点，除去该点后。。。新建图。。然后看所求的最大流有没有减少。。。。来判断每个点是不是割点。。。。然后保存枚举的割点。。就可以了。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define inf 1000000000
#define maxM 500
#define maxN 150000
struct node{
    int u,v,f,next;
}edge[maxN];
int head[maxM],p,lev[maxM],cur[maxM];
int que[maxM];
inline void init1(){
    p=0;memset(head,-1,sizeof(head));
}
bool bfs(int s,int t){
    int qin=0,qout=0,i,u,v;
    memset(lev,0,sizeof(lev));
    lev[s]=1,que[qin++]=s;
    while(qout!=qin){
        u=que[qout++];
        for(i=head[u];i!=-1;i=edge[i].next)
            if(edge[i].f>0 && lev[v=edge[i].v]==0){
                lev[v]=lev[u]+1,que[qin++]=v;
                if(v==t){
                    qout=qin;break;
                }
            }
    }
    return lev[t];
}
int dinic(int s,int t){
    int u,i,f,k,qin;
    int flow=0;
    while(bfs(s,t)){
        memcpy(cur,head,sizeof(head));
        u=s;qin=0;
        while(1){
            if(u==t){
                for(k=0,f=inf;k<qin;k++)
                    if(edge[que[k]].f<f)
                        f=edge[que[i=k]].f;
                for(k=0;k<qin;k++)
                    edge[que[k]].f-=f,edge[que[k]^1].f+=f;
                flow+=f,u=edge[que[qin=i]].u;
            }
            for(i=cur[u];cur[u]!=-1;i=cur[u]=edge[cur[u]].next)
                if(edge[i].f>0 && lev[u]+1==lev[edge[i].v]) break;
            if(cur[u]!=-1)
                que[qin++]=cur[u],u=edge[cur[u]].v;
            else{
                if(qin==0) break;
                lev[u]=-1,u=edge[que[--qin]].u;
            }
        }
    }
    return flow;
}
void addedge(int u,int v,int f){
    edge[p].u=u,edge[p].v=v,edge[p].f=f,edge[p].next=head[u];head[u]=p++;
    edge[p].u=v,edge[p].v=u,edge[p].f=0,edge[p].next=head[v],head[v]=p++;
}
int main(){
    int n,s,t,flow;
    int i,j,vis[maxM],map[500][500],ans[maxM],ain;
    while(scanf("%d%d%d",&n,&s,&t)!=-1){
        init1();ain=0;
        memset(vis,0,sizeof(vis));
        for(i=1;i<=n;i++)
            if(i!=s && i!=t)  addedge(i,i+n,1);
            else  addedge(i,i+n,inf);
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++){
                scanf("%d",&map[i][j]);
                if(map[i][j] && j!=s && i!=t && i!=j) // 这里的条件判断是重点
                    addedge(i+n,j,inf);
            }
        if(map[s][t]){
            printf("NO ANSWER!\n"); continue;
        }
        flow=dinic(s,t+n);
        for(i=1;i<=n;i++){
            if(i==s || i==t) continue;
            vis[i]=1;
            init1();
            for(j=1;j<=n;j++)
                if(vis[j]) continue;
                else if(j!=s && j!=t) addedge(j,j+n,1);
                else addedge(j,j+n,inf);
             for(j=1;j<=n;j++)
                 for(int k=1;k<=n;k++){
                     if(map[j][k] && k!=j && j!=t && k!=s && !vis[j] && !vis[k])//同样是重点
                         addedge(j+n,k,inf);
                 }
                 int tmp=dinic(s,t+n);
                 if(tmp<flow)
                     ans[ain++]=i,flow=tmp;
                 else vis[i]=0;
        }
        printf("%d\n",ain);
        for(i=0;i<ain-1;i++)
            printf("%d ",ans[i]);
        if(ain>0)
            printf("%d\n",ans[ain-1]);
    }
    return 0;
}