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Description
A path with no repeated vertices of an undirected graph is called a simple path. Given an undirected graph and two verteices S and D, return the number of vertics which don‘t lie on any simple paths between S and D.
Input
The input contains multiple test cases.
Each case starts with a line of four integers, N(1 < N ≤ 100), M(1 ≤ M ≤ N(N - 1) / 2), S(0 ≤ S < N), D(0 ≤ D < N). N is the number of vertices, M is the number of edges, S and D are two different vertices. Then M lines follow, each line contains two different integers A(0 ≤ A < N) and B(0 ≤ B < N), which represents an edge of the graph. It‘s ensure that there is at least one simple path between S and D.
Output
Output the number of such vertics, one line per case.
Sample Input
4 3 0 2 0 1 1 2 1 3 4 4 0 2 0 1 1 2 1 3 2 3
Sample Output
1 0
简单路上的点一定与s,d联通
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
int n,m,s,d,mp[105][105],vis[105],mark[105];
void dfs(int x)
{
for(int i=0;i<n;i++)
{
if(!vis[i]&&mp[x][i])
{
vis[i]=1;
dfs(i);
}
}
}
int main()
{
int x,y;
while(scanf("%d%d%d%d",&n,&m,&s,&d)!=EOF)
{
int ans=0;
memset(mp,0,sizeof(mp));
memset(mark,0,sizeof(mark));
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
mp[x][y]=mp[y][x]=1;
}
for(int i=0;i<n;i++)
{
memset(vis,0,sizeof(vis));
vis[i]=1;
if(!vis[s]) dfs(s);
if(!vis[d]) dfs(d);
for(int j=0;j<n;j++)
if(!vis[j]&&j!=s&&j!=d)
mark[j]=1;
}
for(int i=0;i<n;i++)
if(mark[i])
ans++;
printf("%d\n",ans);
}
return 0;
}
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原文地址：http://www.cnblogs.com/a972290869/p/4433352.html
