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

来源：互联网发布：珍玛奇和梁家辉知乎编辑：程序博客网时间：2024/06/05 11:44

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 85using 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;}