11.3.4 例题11-4 UVA 247 Calling Circles （有向图的传递闭包）

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;
}