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 = 1; i <= N; i++)
        for (int j = 1; j <= N; j++)
            if (i != j) dist[i][j] = INF;
            else dist[i][j] = 0;
    for (int i = 0; i < M; i++)
    {
        int mi = 0;
        scanf("%d", &mi);
        for (int j = 0; j < mi; j++)
        {
            scanf("%d", &cow[j]);
        }
        for (int j = 0; j < mi; j++)
        {
            for (int k = 0; k < mi; k++)
            {
                if (cow[j] == cow[k]) continue;
                else dist[cow[j]][cow[k]] = dist[cow[k]][cow[j]] = 1;
            }
        }
    }
    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]);
    for (int i = 1; i <= N; i++)
    {
        sum[i] = 0;
        for (int j = 1; j <= N; j++)
        {
            sum[i] += dist[i][j];
        }
    }
    sort(sum+1, sum+N+1);
    //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[1] / (N-1)*100) << endl;
}