Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands for the bus station that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.
#include<queue> #include<stdio.h> #include<string.h> #define INL 0x3f3f3f3f using namespace std; int vid[10000],x[1080][1080],vist[10000]; int N,M,D; void spfa() { for(int i=0;i<=N;i++) { vist[i]=INL;vid[i]=0; } queue<int> q; vid[0]=1; q.push(0); vist[0]=0; while(!q.empty()) { int u=q.front(); q.pop(); vid[u]=0; for(int i=0;i<=N;i++) { if(vist[i]>vist[u]+x[u][i]) { vist[i]=vist[u]+x[u][i]; if(!vid[i]) { q.push(i); vid[i]=1; } } } } } int main() { while(scanf("%d%d%d",&N,&M,&D)!=EOF) { memset(x,INL,sizeof(x)); int a,b,c; for(int i=0;i<M;i++) { scanf("%d%d%d",&a,&b,&c); if(x[a][b]>c) x[a][b]=c; } int n,g,min; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&g); x[0][g]=0;//万能源点知识 } spfa(); if(vist[D]==INL) printf("-1\n"); else printf("%d\n",vist[D]); } return 0; }