Choose the best route

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 6

Problem Description

One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.

Input

There are several test cases. Each case begins with three integers n, m and s,(n<1000,m<20000,1=<S<=N) q n in this city directed .(Maybe are several ways between two .) s bus that near Kiki’s friend’s home. follow m lines ,each contains three , (0<t<="1000)." from p to station there is way and it will costs t minutes . a line with an integer w(0<w<n), means number of stations Kiki can take at the beginning. Then follows w integers stands for these stations. div <>

Output

The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.

Sample Input

5 8 5 1 2 2 1 5 3 1 3 4 2 4 7 2 5 6 2 3 5 3 5 1 4 5 1 2 2 3 4 3 4 1 2 3 1 3 4 2 3 2 1 1

Sample Output

1 -1

题解：把终点当起点做

代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define MIN(x,y)(x<y?x:y)
 4 const int MAXN=1010;
 5 const int INF=0x3f3f3f3f;
 6 int n,vis[MAXN],d[MAXN];
 7 int map[MAXN][MAXN];
 8 void initial(){
 9     memset(map,0x3f,sizeof(map));
10     memset(vis,0,sizeof(vis));
11     memset(d,INF,sizeof(d));
12 }
13 void dijskra(int s){
14     d[s]=0;
15     while(true){
16         int k=-1;
17         for(int i=1;i<=n;i++)
18             if(!vis[i]&&(k==-1||d[i]<d[k]))k=i;
19         if(k==-1)break;
20         vis[k]=1;
21         for(int i=1;i<=n;i++)
22             d[i]=MIN(d[i],d[k]+map[k][i]);
23     }
24 }
25 int main(){
26     int m,s,p,q,t,a,e,ans;
27     while(~scanf("%d%d%d",&n,&m,&s)){
28             initial();
29         while(m--){
30             scanf("%d%d%d",&p,&q,&t);
31             if(map[q][p]>t)map[q][p]=t;
32         }
33         scanf("%d",&a);
34         ans=INF;
35         dijskra(s);
36         while(a--){
37             scanf("%d",&e);
38             ans=MIN(ans,d[e]);
39         }
40         if(ans==INF)puts("-1");
41         else printf("%d\n",ans);
42     }
43     return 0;
44 }