任意两点间最短距离floyd-warshall ---- POJ 2139 Six Degrees of Cowvin Bacon

floyd-warshall算法 通过dp思想 求任意两点之间最短距离

重复利用数组实现方式dist[i][j] i - j的最短距离

for(int k = 1; k <= N; k++)

for (int i = 1; i <= N; i++)

　　for (int j = 1; j <= N; j++)

　　　　dist[i][j] = min(dist[i][j], dist[i][k]+dist[k][j]);

非常好实现 O(V^3)

这里贴一道刚好用到的题

http://poj.org/problem?id=2139

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 #define INF 0x3f3f3f3f
 #define MAXV 307
 using namespace std;

 //floyd-warshall

 int dist[MAXV][MAXV];
 int main()
 {
     int N, M;
     int cow[MAXV];
     double sum[MAXV];
     freopen("in.txt", "r", stdin);
     scanf("%d%d", &N, &M);
     for (int i = ; i <= N; i++)
         for (int j = ; j <= N; j++)
             if (i != j) dist[i][j] = INF;
             else dist[i][j] = ;
     for (int i = ; i < M; i++)
     {
         int mi = ;
         scanf("%d", &mi);
         for (int j = ; j < mi; j++)
         {
             scanf("%d", &cow[j]);
         }
         for (int j = ; j < mi; j++)
         {
             for (int k = ; k < mi; k++)
             {
                 if (cow[j] == cow[k]) continue;
                 else dist[cow[j]][cow[k]] = dist[cow[k]][cow[j]] = ;
             }
         }
     }
     for (int k = ; k <= N; k++)
         for (int i = ; i <= N; i++)
             for (int j = ; j <= N; j++) dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
     for (int i = ; i <= N; i++)
     {
         sum[i] = ;
         for (int j = ; j <= N; j++)
         {
             sum[i] += dist[i][j];
         }
     }
     sort(sum+, sum+N+);
     //注意题目最后的要求 A single integer that is 100 times the shortest mean degree of separation of any of the cows.
     //所以要变为int
     cout << int(sum[] / (N-)*) << endl;
 }