UVa 821 网页跳跃（Floyd）

https://vjudge.net/problem/UVA-821

题意：
给出一个有向图，任意两点都可相互到达，求任意两点的最短距离的平均值。

思路：
求两点的最短距离，用Floyd算法很方便，最后加起来算个平均值即可。

#include<iostream> 
#include<algorithm>
#include<cstring>
using namespace std;

const int INF = 1000;

int x, y;
int d[105][105];

void Floyd()
{
    for (int k = 1; k <= 100;k++)
    for (int i = 1; i <= 100;i++)
    for (int j = 1; j <= 100; j++)
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    int kase = 0;
    while (cin >> x >> y && (x!=0 || y!=0))
    {
        for (int i = 1; i <= 100;i++)
        for (int j = 1; j <= 100;j++)
        if (i == j)  d[i][j] = 0;
        else d[i][j] = INF;

        do
        {
            d[x][y] = 1;
        } while (cin >> x >> y && (x!=0 || y!=0));
        Floyd();
        double ans = 0;
        int count = 0;
        for (int i = 1; i <= 100;i++)
        for (int j = 1; j <= 100; j++)
        {
            if (i!=j && d[i][j] < INF)
            {
                count++;
                ans += d[i][j];
            }
        }
        printf("Case %d: average length between pages = %.3f clicks
", ++kase, ans / count);
    }
}