HDOJ 2807 - The Shortest Path 读题细节+矩阵相乘+Floyd

     注意题目的一个重要细节...A*B=C..并且A,B,C分别是三个不同的城市

     矩阵相乘构造图时..注意..枚举了当前A,B后..先算出矩阵.再来找C....

     最后Floyd跑出答案...

     我写的很暴力了...对于矩阵比较这一块..看别人的题解...有更好的方法...用把每个矩阵*向量(1,2,3,4...m)变成一个1维的向量..再进行比较...

Program:

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<queue>
#include<stack>
#include<set>
#include<algorithm>
#define ll long long
#define oo 1000000007
#define pi acos(-1.0)
#define MAXN 85
using namespace std;   
struct node
{
       int s[MAXN][MAXN];
}City[MAXN];
int N,M;
int arc[MAXN][MAXN];
node mul(node a,node b)
{
       int i,j,k;
       node h;
       memset(h.s,0,sizeof(h.s));
       for (k=1;k<=M;k++)
          for (i=1;i<=M;i++)
             for (j=1;j<=M;j++) 
                h.s[i][j]+=a.s[i][k]*b.s[k][j]; 
       return h;
}
void Floyd()
{
       int k,i,j;
       for (k=1;k<=N;k++)
          for (i=1;i<=N;i++)
             for (j=1;j<=N;j++)
                arc[i][j]=min(arc[i][j],arc[i][k]+arc[k][j]);
       return;
}
int main()
{   
       int i,t,j,x,k;
       node p;  
       while (~scanf("%d%d",&N,&M) && N && M)
       {
              for (t=1;t<=N;t++)
                 for (i=1;i<=M;i++)
                    for (j=1;j<=M;j++)
                       scanf("%d",&City[t].s[i][j]); 
              memset(arc,0x3f,sizeof(arc)); 
              for (i=1;i<=N;i++) arc[i][i]=0;
              for (t=1;t<=N;t++)
                 for (i=1;i<=N;i++)
                 if (i!=t)
                 { 
                       p=mul(City[i],City[t]);
                       for (j=1;j<=N;j++) 
                       {
                             if (j==t || j==i) continue;
                             for (x=1;x<=M;x++)
                               for (k=1;k<=M;k++)
                                   if (p.s[x][k]!=City[j].s[x][k]) goto A;
                             arc[i][j]=1;
                             A: ;
                       }
                 }
              Floyd();
              scanf("%d",&t);
              while (t--)
              {
                     scanf("%d%d",&i,&j);
                     if (arc[i][j]>oo) printf("Sorry\n"); 
                         else printf("%d\n",arc[i][j]);
              }
       } 
       return 0;
}