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题目大意：
给你n个人，m条边，a->b,b->a，才能说这两个人是联通的。问现在有多少个联通圈。输出每个联通圈。n<=25
解题思路：
直接建图，用有向图的闭包传递特性来处理每个人之间的联通关系。然后dfs一次，遍历邻接矩阵中与某个点相连的几个点，边遍历， 边打印输出。
代码：
# include<cstdio>
# include<iostream>
# include<map>
# include<cstring>
# include<vector>
# include<set>
using namespace std;
# define MAX 55
map<string,int>MP;
vector<string>name;
int e[MAX][MAX];
int vis[MAX];
int n,m;
int change ( string s )
{
for ( int i = 0;i < name.size();i++ )
{
if ( name[i]==s )
return i;
}
name.push_back(s);
return name.size()-1;
}
void dfs( int u )
{
vis[u] = 1;
for ( int v = 0;v < n;v++ )
{
if ( vis[v]==0&&e[u][v]&&e[v][u] )
{
cout<<","<<" "<<name[v];
dfs(v);
}
}
return;
}
int main(void)
{
int icase = 1;
while ( scanf("%d%d",&n,&m)!=EOF )
{
if ( n==0&&m==0 )
break;
memset(e,0,sizeof(e));
name.clear();
while ( m-- )
{
string s1,s2; cin>>s1>>s2;
int t1 = change(s1);
int t2 = change(s2);
e[t1][t2] = 1;
}
for ( int i = 0;i < n;i++ )
e[i][i] = 1;
for ( int k = 0;k < n;k++ )
{
for ( int i = 0;i < n;i++ )
{
for ( int j = 0;j < n;j++ )
{
e[i][j] = e[i][j]||(e[i][k]&&e[k][j]);
}
}
}
if ( icase )
puts("");
printf("Calling circles for data set %d:\n",icase++);
memset(vis,0,sizeof(vis));
for ( int i = 0;i < n;i++ )
{
if (vis[i]==0)
{
cout<<name[i];
dfs(i);
puts("");
}
}
}
return 0;
}
11.3.4 例题11-4 UVA 247 Calling Circles （有向图的传递闭包）
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原文地址：http://www.cnblogs.com/lushichuanshuo/p/4856599.html
