题目链接：

option=com_onlinejudge&Itemid=8&page=show_problem&problem=508">https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=508

题意：20个点的随意最短路。floyd

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <string>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iostream>
#include <ctime>
#include <cmath>
#include <cstring>

using namespace std;

int a[25][25];
char c;

int main()
{
    int cases=1;
    int n,m;
    while (~scanf("%d",&n))
    {
        for(int i=1;i<=20;i++)
                for(int j= 1;j<=20;j++)
                {
                    if (i==j) a[i][j] = 0;
                    else a[i][j] = 100000000;
                }
        while(n--)
            {
                scanf("%d",&m);
                a[1][m] = a[m][1] = 1;
            }
        for(int i=2;i<=19;i++)
        {
            scanf("%d",&n);
            while(n--)
            {
                scanf("%d",&m);
                a[i][m] = a[m][i] = 1;
            }
        }
        for(int k=1;k<=20;k++)
            for(int i=1;i<=20;i++)
                for(int j= 1;j<=20;j++)
        {
            if (a[i][k] + a[k][j] < a[i][j])
                a[i][j] =  a[i][k] + a[k][j];
        }
        int s,d;
        scanf("%d",&n);
        printf("Test Set #%d\n",cases++);
        while(n--)
        {
            scanf("%d%d",&s,&d);
            printf("%2d to %2d: %d\n",s,d,a[s][d]);
        }
        printf("\n");
    }
    return 0;
}