Friendship （poj 1815 最小点割集+枚举）

Language:Default

Friendship

Time Limit: 2000MS Memory Limit: 20000KTotal Submissions: 9930 Accepted: 2743

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

Source

POJ Monthly

题意：给出N个人之间的通讯关系，给出S和T，问切断S到T的联系最小要删去的点是多少，输出最小字典序解。

思路：拆点，从1~N枚举删掉i后看最大流是否减少，若减少则i属于最小点割集。

代码：

#include <iostream>#include <functional>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,r#define FRE(i,a,b)  for(i = a; i <= b; i++)#define FREE(i,a,b) for(i = a; i >= b; i--)#define FRL(i,a,b)  for(i = a; i < b; i++)#define FRLL(i,a,b) for(i = a; i > b; i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define DBG         pf("Hi\n")typedef long long ll;using namespace std;#define INF 0x3f3f3f3f#define mod 1000000009const int maxn = 500;const int MAXN = 500;const int MAXM = 200010;int N,S,T;int a[maxn][maxn];struct Edge{    int to,next,cap,flow;}edge[MAXM];int tol;int head[MAXN];int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];bool Remove[maxn];void init(){    tol=0;    memset(head,-1,sizeof(head));}//加边，单向图三个参数，双向图四个参数void addedge(int u,int v,int w,int rw=0){    edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];    edge[tol].flow=0; head[u]=tol++;    edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];    edge[tol].flow=0; head[v]=tol++;}//输入参数：起点，终点，点的总数//点的编号没有影响，只要输入点的总数int sap(int start,int end,int N){    memset(gap,0,sizeof(gap));    memset(dep,0,sizeof(dep));    memcpy(cur,head,sizeof(head));    int u=start;    pre[u]=-1;    gap[0]=N;    int ans=0;    while (dep[start]<N)    {        if (u==end)        {            int Min=INF;            for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])                if (Min>edge[i].cap-edge[i].flow)                    Min=edge[i].cap-edge[i].flow;            for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])            {                edge[i].flow+=Min;                edge[i^1].flow-=Min;            }            u=start;            ans+=Min;            continue;        }        bool flag=false;        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=true;                cur[u]=pre[v]=i;                break;            }        }        if (flag)        {            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;            }        gap[dep[u]]--;        if (!gap[dep[u]]) return ans;        dep[u]=Min+1;        gap[dep[u]]++;        if (u!=start) u=edge[pre[u]^1].to;    }    return ans;}void Build(){    init();    for (int i=1;i<=N;i++)    {        if (!Remove[i])            addedge(i,i+N,1);        for (int j=1;j<=N;j++)        {            if (a[i][j]&&i!=j)                addedge(i+N,j,INF);        }    }}int main(){#ifndef ONLINE_JUDGE    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);#endif    int i,j,x;    while (~scanf("%d%d%d",&N,&S,&T))    {        bool flag=true;        for (i=1;i<=N;i++)            for (j=1;j<=N;j++)            {                scanf("%d",&a[i][j]);                if (a[i][j]&&((i==S&&j==T)||(i==T&&j==S)))                    flag=false;            }        if (!flag){            printf("NO ANSWER!\n");            continue;        }        memset(Remove,false,sizeof(Remove));        vector<int>out;        out.clear();        S=S+N;        Build();        int ans=sap(S,T,2*N);        printf("%d\n",ans);        for (i=1;i<=N;i++)        {            if (i==S-N||i==T) continue;            Remove[i]=true;            Build();            if (sap(S,T,2*N)<ans)            {                ans--;                out.push_back(i);            }            else Remove[i]=false;            if (ans<=0) break;        }        if (out.size()>0)        {            printf("%d",out[0]);            for (i=1;i<out.size();i++)                printf(" %d",out[i]);            printf("\n");        }    }    return 0;}