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POJ 1815 最小點割集

編輯：C++入門知識

Friendship Time Limit: 2000MS Memory Limit: 20000K Total 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

Sample Output

1
2

題意：給定一個圖，求去掉最少的頂點，使得s與t不連通，答案有多個時，按輸出字典序最小的。

假如s與t直接連通，則無解，

否則拆點，一個點拆成入度，出度兩個點，之間s，t兩個點流量為無窮大，其他點為1，然後建圖，跑一遍最大流，

然後從小到大枚舉除s，t之外的點，用一個數組標記把點去掉，建圖跑最大流，假如流量減少，則說明這個點需要去掉，否則這個點無關緊要，不能去掉，

然後輸出答案即可。

代碼：

/* ***********************************************
Author :rabbit
Created Time :2014/3/8 16:40:10
File Name :1718.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
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]edge[S[i]].cap-edge[S[i]].flow)
                    MIN=edge[S[i]].cap-edge[S[i]].flow,id=i;
            for(int i=0;i 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