HDOJ Choose the best route

 

Choose the best route

http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=12497&pid=1011

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

Total Submission(s) : 9   Accepted Submission(s) : 3

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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) 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.

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

Author

dandelion

Source

2009浙江大学计算机研考复试（机试部分）——全真模拟

 

一开始这题看的不仔细，习惯性的就把代码敲完了，一次性测试数据正确却通不过，靠，原来被题目忽悠了，输入p q t 的时候是从q从到p， 我搞了半天都是从p到q，原来啊

 

代码如下：

 

#include<iostream>
using namespace std;
const int maxnum=1005;
const int maxint=9999999;
int dist[maxnum];
int map[maxnum][maxnum];
bool s[maxnum];
int n,m,t; //分别是车站数，道路数，始点
void init(){
      for(int i=1;i<=n;i++)
         for(int j=1;j<=n;j++) 
              map[i][j]=maxint;
}
void Dijkstra(int v){
     for(int i=1;i<=n;i++){
             dist[i]=map[v][i];
             s[i]=0;
     }
     s[v]=1;
     dist[v]=0;
     for(int i=2;i<=n;i++){
            int temp=maxint;
            int u=v;
           for(int j=1;j<=n;j++)
                if((!s[j]) && dist[j]<temp){
                           temp=dist[j];
                           u=j;
                }
           s[u]=1;
           for(int j=1;j<=n;j++)
               if((!s[j]) &&  dist[u]+map[u][j]<dist[j])
                  dist[j]=dist[u]+map[u][j];
     }
}
int main(){
    int a,b,d;
    while(scanf("%d%d%d",&n,&m,&t)!=EOF){
          init();
          for(int i=1;i<=m;i++){
                  scanf("%d%d%d",&a,&b,&d); //注意这里是从b 到 a
                  if(d<map[b][a])
                      map[b][a]=d;  //注意这里是有向图
          }
      
          Dijkstra(t);
          int w,min=maxint,x;
          scanf("%d",&w);  
          for(int i=1;i<=w;i++){
                  scanf("%d",&x);
                  if(dist[x]<min)
                  min=dist[x];
          }
          if(min<maxint)
          printf("%d\n",min);
          else
          printf("-1\n");                             
    }
return 0;
}