本文主要是介绍hdu 2807 The Shortest Path（矩阵相乘+floyd），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
http://acm.hdu.edu.cn/showproblem.php?pid=2807
大致题意：给出n个m*m的矩阵，若存在三个互异的矩阵满足a*b = c，那么a,c之间存在权值为1的单向边。有询问u,v，输出uv之间的最短距离。
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#define LL long long
#define _LL __int64
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 85;int n,m;
int a[maxn][maxn][maxn];
int Map[maxn][maxn];void init()
{for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(i == j) Map[i][j] = 0;else Map[i][j] = INF;}}for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){for(int k = 1; k <= m; k++)scanf("%d",&a[i][j][k]);}}for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(i == j) continue;//任意两个矩阵相乘int tmp[maxn][maxn];memset(tmp,0,sizeof(tmp));for(int x = 1; x <= m; x++){for(int y = 1; y <= m; y++){for(int z = 1; z <= m; z++)tmp[x][y] += a[i][x][z] * a[j][z][y];}}//枚举与tmp相等的矩阵for(int k = 1; k <= n; k++){if(k == i || k == j) continue;int flag = 1;for(int x = 1; x <= m; x++){for(int y = 1; y <= m; y++){if(a[k][x][y] != tmp[x][y]){flag = 0;break;}}if(!flag) break;}if(flag)Map[i][k] = 1;}}}
}void floyd()
{for(int k = 1; k <= n; k++){for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(Map[i][j] > Map[i][k] + Map[k][j])Map[i][j] = Map[i][k] + Map[k][j];}}}
}int main()
{while(~scanf("%d %d",&n,&m)){if(n == 0 && m == 0) break;init();floyd();int k;scanf("%d",&k);while(k--){int u,v;scanf("%d %d",&u,&v);if(Map[u][v] == INF)printf("Sorry\n");else printf("%d\n",Map[u][v]);}}return 0;
}
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