POJ 1815 最小点割集

Friendship

Time Limit: 2000MS Memory Limit: 20000KTotal Submissions: 8640 Accepted: 2412

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 31 1 01 1 10 1 1

Sample Output

12

题意：给定一个图，求去掉最少的顶点，使得s与t不连通，答案有多个时，按输出字典序最小的。

假如s与t直接连通，则无解，

否则拆点，一个点拆成入度，出度两个点，之间s，t两个点流量为无穷大，其他点为1，然后建图，跑一遍最大流，

然后从小到大枚举除s，t之外的点，用一个数组标记把点去掉，建图跑最大流，假如流量减少，则说明这个点需要去掉，否则这个点无关紧要，不能去掉，

然后输出答案即可。

代码：

/* ***********************************************Author :rabbitCreated Time :2014/3/8 16:40:10File Name :1718.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;const int maxn=1010;const int maxm=400010;struct Edge{    int to,next,cap,flow;    Edge(){};    Edge(int _next,int _to,int _cap,int _flow){        next=_next;to=_to;cap=_cap;flow=_flow;    }}edge[maxm];int head[maxn],tol,gap[maxn],dep[maxn],cur[maxn];void addedge(int u,int v,int flow){    edge[tol]=Edge(head[u],v,flow,0);head[u]=tol++;    edge[tol]=Edge(head[v],u,0,0);head[v]=tol++;}int Q[maxn];void bfs(int start,int end){    memset(dep,-1,sizeof(dep));    memset(gap,0,sizeof(gap));    gap[0]++;int front=0,rear=0;    dep[end]=0;Q[rear++]=end;    while(front!=rear){        int u=Q[front++];        for(int i=head[u];i!=-1;i=edge[i].next){            int v=edge[i].to;if(dep[v]==-1&&edge[i].cap)                Q[rear++]=v,dep[v]=dep[u]+1,gap[dep[v]]++;        }    }}int S[maxn];int sap(int start,int end,int N){    bfs(start,end);    memcpy(cur,head,sizeof(head));    int top=0,u=start,ans=0;    while(dep[start]<N){        if(u==end){            int MIN=INF,id;            for(int i=0;i<top;i++)                if(MIN>edge[S[i]].cap-edge[S[i]].flow)                    MIN=edge[S[i]].cap-edge[S[i]].flow,id=i;            for(int i=0;i<top;i++)                edge[S[i]].flow+=MIN,edge[S[i]^1].flow-=MIN;            ans+=MIN,top=id,u=edge[S[top]^1].to;            continue;        }        bool flag=0;int v;        for(int i=cur[u];i!=-1;i=edge[i].next){            v=edge[i].to;            if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u]){                flag=1;cur[u]=i;break;            }        }        if(flag){            S[top++]=cur[u];u=v;continue;        }        int MIN=N;        for(int i=head[u];i!=-1;i=edge[i].next)            if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<MIN)                MIN=dep[edge[i].to],cur[u]=i;        if(--gap[dep[u]]==0)break;gap[dep[u]=MIN+1]++;        if(u!=start)u=edge[S[--top]^1].to;    }    return ans;}int flag[400][400],mark[500];int main(){     //freopen("data.in","r",stdin);     //freopen("data.out","w",stdout);     int n,s,t; while(~scanf("%d%d%d",&n,&s,&t)){ memset(head,-1,sizeof(head));tol=0; memset(mark,0,sizeof(mark)); for(int i=1;i<=n;i++){ if(i==s||i==t)addedge(i,i+n,INF); else addedge(i,i+n,1); for(int j=1;j<=n;j++){ scanf("%d",&flag[i][j]); if(flag[i][j]&&i!=j) addedge(i+n,j,INF); } } if(flag[s][t]){              printf("NO ANSWER!\n");              continue;  } vector<int> ans; addedge(0,s,INF);addedge(t+n,2*n+1,INF); int pre=sap(0,2*n+1,4*n+10); for(int i=1;i<=n;i++){ if(i==s||i==t)continue; mark[i]=1; memset(head,-1,sizeof(head));tol=0; for(int j=1;j<=n;j++){ if(mark[j])continue; if(j!=s&&j!=t)addedge(j,j+n,1); else addedge(j,j+n,INF); } for(int j=1;j<=n;j++){ for(int k=1;k<=n;k++){ if(flag[j][k]&&k!=j&&!mark[j]&&!mark[k]) addedge(j+n,k,INF); } } addedge(0,s,INF);addedge(t+n,2*n+1,INF); int ret=sap(0,2*n+1,4*n+10); if(ret<pre)ans.push_back(i),pre=ret; else mark[i]=0; } printf("%d\n",ans.size()); if(ans.size()){ for(int i=0;i<ans.size();i++) printf("%d%c",ans[i],i==ans.size()-1?'\n':' '); } }     return 0;}