模板题，求一个点到任何一点的距离，用Floyd就行了，结点不一定是从1 ~ n 的，所以需要记录结点的id

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<list>
#include<cmath>
#include<string>
#include<sstream>
#include<ctime>
using namespace std;
#define _PI acos(-1.0)
#define INF (1 << 10)
#define esp 1e-9
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pill;
/*===========================================
===========================================*/
#define MAXD 100 + 10
int G[MAXD][MAXD];
int main(){
    int a , b;
    int Case = 1;
    while(scanf("%d%d",&a,&b)){
        if(!a && !b) break;
        for(int i = 1 ; i <= 100 ; i++)
            for(int j = 1; j <= 100 ; j++)
                if(i == j) G[i][j] = 0;
                else       G[i][j] = INF;
        map<int,int>vis;
        vector<int>node;
        G[a][b] = 1;
        while(scanf("%d%d",&a,&b)){
            if(!a && !b) break;
            G[a][b] = 1;
            if(!vis.count(a)){
                node.push_back(a);
                vis[a] = 1;
            }
            if(!vis.count(b)){
                node.push_back(b);
                vis[b] = 1;
            }
        }
        for(int k = 0 ; k < node.size() ; k++)
            for(int i = 0 ; i < node.size() ; i++)
                for(int j = 0 ; j < node.size() ; j++)
                    if(G[node[i]][node[k]] < INF && G[node[k]][node[j]] < INF)
                        G[node[i]][node[j]] = min(G[node[i]][node[j]],G[node[i]][node[k]] + G[node[k]][node[j]]);
        int sum = 0;
        for(int i = 0 ; i < node.size() ; i++)
            for(int j = 0 ; j < node.size() ; j++){
                sum += G[node[i]][node[j]];
            }
        double ans = 1.0 * sum / ((node.size() -1)* node.size());
        printf("Case %d: average length between pages = %.3f clicks\n",Case++,ans);
    }
    return 0;
}