UVA Page Hopping（Floyd算法）

题意：

给出若干个点， 求出每对点之间距离的最小平均值。

解题思路：

直接Floyd求出所有点两两之间最短路径累和即可。

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=762

Memory: 0 KB		Time: 26 MS
Language: C++ 4.8.2		Result: Accepted

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>

using namespace std;

#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)
#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)

int buf[10];
inline long long read()
{
    long long x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

inline void writenum(int i)
{
    int p = 0;
    if(i == 0) p++;
    else while(i)
        {
            buf[p++] = i % 10;
            i /= 10;
        }
    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);
}
/**************************************************************/
#define MAX_N 110
const int INF = 0x3f3f3f3f;
int G[MAX_N][MAX_N];
map<int, int> mp;
int main()
{
    int n, m;
    int cas = 1;
    while(~scanf("%d%d", &n, &m) && (n + m))
    {
        memset(G, INF, sizeof(G));
        for(int i = 0 ; i < MAX_N ; i++)
            G[i][i] = 0;
        mp.clear();
        int t = 0;
        if(!mp.count(n))
        {
            mp[n] = t++;
        }
        if(!mp.count(m))
        {
            mp[m] = t++;
        }
        int u = mp[n];
        int v = mp[m];
        G[u][v] = 1;

        while(scanf("%d%d", &n, &m) && (n + m))
        {
            if(!mp.count(n))
            {
                mp[n] = t++;
            }
            if(!mp.count(m))
            {
                mp[m] = t++;
            }
            int u = mp[n];
            int v = mp[m];
            G[u][v] = 1;
        }
        int cnt = mp.size();
        int sum = 0;
        for(int k = 0 ; k < cnt ; k++)
            for(int i = 0 ; i < cnt ; i++)
                for(int j = 0 ; j < cnt ; j++)
                {
                    G[i][j] = min(G[i][j], G[i][k] + G[k][j]);
                }
        for(int i = 0 ; i < cnt ; i++)
            for(int j = 0 ; j < cnt ; j++)
                {
                    sum += G[i][j];
                }
        double ans = (sum + 0.0)/ (cnt * (cnt - 1));
        printf("Case %d: average length between pages = %.3f clicks\n", cas++, ans);
    }

    return 0;
}