poj 1135 Domino Effect 单源最短…

单源最短路径的一个变形

解法：先求出单源最短路

ans[i][j]表示i到j这条边上最后倒下的domino骨牌倒下的时间

ans[i][j]=(dist[i]+dist[j]+e[i][j])/2.0

ps：在双向边上wa了，并且在边的初始化上wa了，非常2的把边的初始值定为了0，并且计算ans[i][j]时没有对边是否存在做判断

#include <iostream>
 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 using namespace std;
 const int maxn=501;
 const int inf=1<<30;
 int n,m;
 int e[maxn][maxn],dist[maxn];
 float answer[maxn][maxn];

int dijkstra(int n)
 {
     int txt[maxn];
     memset(txt,0,sizeof(txt));
     for(int i=1;i<=n;i++)
     dist[i]=inf;
     dist[1]=0;
     for(int k=2;k<=n;k++)
     {
         int ted=inf,tmp=n+1;
         for(int i=1;i<=n;i++)
         if(txt[i]==0&&dist[i]<ted)
         {
             ted=dist[i];
             tmp=i;
         }
         if(tmp==n+1) return(0);
         txt[tmp]=1;
         for(int i=1;i<=n;i++)
         if(e[tmp][i]!=inf&&dist[tmp]+e[tmp][i]<dist[i])
         dist[i]=dist[tmp]+e[tmp][i];
     }
     return 0;
 }

int main()
 {
     int tcase=0;
     while(++tcase)
     {
         scanf("%d %d",&n,&m); if(n==0&&m==0) break;
         for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++)
         e[i][j]=inf;
         for(int i=1;i<=n;i++) e[i][i]=0;

        for(int a,b,c,i=1;i<=m;i++)
         {
             scanf("%d %d %d",&a,&b,&c);
             e[a][b]=c;
             e[b][a]=c;
         }

        dijkstra(n);

        memset(answer,0,sizeof(answer));
         for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++)
         if(e[i][j]!=inf)
         answer[i][j]=(dist[i]+dist[j]+e[i][j])/2.0;
         float ans=0;
         for(int i=1;i<=n;i++) if(ans<dist[i]) ans=dist[i];
         int ture=0;
         for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++)
         if(answer[i][j]>ans)
         {
             ans=answer[i][j];
             ture=1;
         }
         printf("System #%d\n",tcase);
 //        printf("%d\n",ture);
         if(ture==0)
         {
             for(int i=1;i<=n;i++)
             if(dist[i]==ans)
             printf("The last domino falls after %.1f seconds, at key domino %d.\n",ans,i);
         }
         else
         {
             for(int i=1;i<=n;i++)
             for(int j=1;j<=n;j++)
             if(answer[i][j]==ans)
             {
                 printf("The last domino falls after %.1f seconds, between key dominoes %d and %d.\n",ans,i,j);
                 i=n+1;
                 break;
             }
         }
         printf("\n");
     }
     return 0;
 }