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Friendship
| Time Limit: 2000MS |
|
Memory Limit: 20000K |
| Total Submissions: 8747 |
|
Accepted: 2451 |
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
Sample Input
3 1 3
1 1 0
1 1 1
0 1 1
Sample Output
1
2
Source
POJ Monthly
给你一个图，要求至少删除图中几个点，使得起点到终点不连通。如果有多种方案，输出序号最下的那种。
最小点割集：无向(有向)图G中，给定源点s和终点t，至少要删去多少个点(具体一点，删哪些点)，使得s和t不连通。
解法：一般最小点割转化到最小边割上，将原图中的点v拆成v'和v''，且w(v‘,v'')=1。对于原图中的有向边(u,v)，则有w(u'',v')=INF；若是无向边，则还要加上边：w(v',u'')=INF。然后求以s''为源点，t'为汇点的最大流。maxflow即为最少需要删的点数，割边集对应了具体删的点的一组解。
题目要求输出序号最小的，我们可以按照序号从小到大枚举每一个点，然后删除i'和i''，如果新的最大流比原来的最大流小，说明这个点在最小割里面，否则回复删除的边。
//1408K 704MS
#include
#include
#include
#define inf 0x3f3f3f3f
#define M 1007
#define MIN(a,b) a>b?b:a;
using namespace std;
struct E
{
int v,w,next;
} edg[500000];
int dis[M],gap[M],head[M],nodes;
int sourse,sink,nn;
int g[M][M],cut[M],p[M];
int n;
void addedge(int u,int v,int w)
{
edg[nodes].v=v;
edg[nodes].w=w;
edg[nodes].next=head[u];
head[u]=nodes++;
edg[nodes].v=u;
edg[nodes].w=0;
edg[nodes].next=head[v];
head[v]=nodes++;
}
int dfs(int src,int aug)
{
if(src==sink)return aug;
int left=aug,mindis=nn;
for(int j=head[src]; j!=-1; j=edg[j].next)
{
int v=edg[j].v;
if(edg[j].w)
{
if(dis[v]+1==dis[src])
{
int minn=MIN(left,edg[j].w);
minn=dfs(v,minn);
edg[j].w-=minn;
edg[j^1].w+=minn;
left-=minn;
if(dis[sourse]>=nn)return aug-left;
if(left==0)break;
}
if(dis[v]
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