UVa 247 电话圈（Floyd传递闭包）

https://vjudge.net/problem/UVA-247

题意：

如果两个人相互打电话，则说他们在同一个电话圈里。例如，a打给b，b打给c，c打给d，d打给a，则这4个人在同一个圈里；如果e打给f但f不打给e，则不能推出e和f在同一个电话圈里，输出所有电话圈。

思路：

通过Floyd求一个传递闭包。最后dfs输出每一个电话圈即可。

传递闭包的求法：

for (int k = 0; k < n;k++)
for (int i = 0; i < n;i++)
for (int j = 0; j < n; j++)
    d[i][j] = d[i][j] || (d[i][k] && d[k][j]);

#include<iostream> 
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;

int n, m;
int d[30][30];
string s1, s2;
map<string, int> ID;
vector<string> name;
int vis[30];

void dfs(int k)
{
    vis[k] = 1;
    for (int i = 0; i < n; i++)
    {
        if (d[k][i] && d[i][k])
        {
            if (!vis[i])
            {
                cout << ", " << name[i];
                dfs(i);
            }
        }
    }
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int kase = 1;
    while (scanf("%d%d", &n, &m))
    {
        if (n == 0 && m == 0)    break;
        if (kase > 1)     printf("
");
        memset(d, 0, sizeof(d));
        memset(vis, 0, sizeof(vis));
        ID.clear();
        name.clear();
        int cnt = 0;
        for (int i = 0; i < m; i++)
        {
            cin >> s1 >> s2;
            if (!ID.count(s1))
            {
                ID[s1] = cnt++;
                name.push_back(s1);
            }
            if (!ID.count(s2))
            {
                ID[s2] = cnt++;
                name.push_back(s2);
            }
            int x = ID[s1];
            int y = ID[s2];
            d[x][y] = 1;
        }

        for (int k = 0; k < n;k++)
        for (int i = 0; i < n;i++)
        for (int j = 0; j < n; j++)
            d[i][j] = d[i][j] || (d[i][k] && d[k][j]);

        printf("Calling circles for data set %d:
", kase++);
        for (int i = 0; i < n; i++)
        {
            if (!vis[i])
            {
                cout << name[i];
                dfs(i);
                printf("
");
            }
        }
    }
    return 0;
}