HDU-2066(dijkstra最短路综合)

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此题目,就是一定要注意建双路边,而且因为不知道有多少个顶点,所以开始的时候就要建最多的情况.也就是1005(习惯性的多出几个咯),然后就是一个起点,对应的多个终点的两重循环就好,其实还是比较简单的.

贴出代码:

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#define inf 0x3fffffffint M,S,D;//M represents the number of edges,S stands for the number of the beginning point,and D represents the number of destination;int visit[1005];int dis[1005];int map[1005][1005];int sta[1005];int des[1005];int dijkstra(int st){for(int i=1;i<=1005;i++)dis[i]=inf;dis[st]=0;for(int j=1;j<=1005;j++){int t=inf,pos;for(i=1;i<=1005;i++){if(!visit[i]&&t>dis[i]){t=dis[i];pos=i;}}visit[pos]=i;for(i=1;i<=1005;i++){if(!visit[i]&&dis[i]>dis[pos]+map[pos][i]&&map[pos][i]!=0x3f3f3f3f){dis[i]=dis[pos]+map[pos][i];}}}int min=inf;for(int k=1;k<=D;k++){if(min>dis[des[k]]){min=dis[des[k]];}}return min;}int main(){while(scanf("%d%d%d",&M,&S,&D)!=EOF){int a,b,time;memset(map,0x3f,sizeof(map));for(int i=1;i<=M;i++){scanf("%d%d%d",&a,&b,&time);if(time<map[a][b]){map[a][b]=time;map[b][a]=time;}}for(int j=1;j<=S;j++){scanf("%d",&sta[j]);}for(i=1;i<=D;i++){scanf("%d",&des[i]);}int t=inf;for(j=1;j<=S;j++){memset(visit,0,sizeof(visit));int ans=dijkstra(sta[j]);if(t>ans){t=ans;}}printf("%d\n",t);}return 0;}