Choose the best route--hdu2680

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10482    Accepted Submission(s): 3373

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

 

 

 

很明显的要用dijkstra，但是估计大家用一般的方法建图的时候都会超时吧，因为起点不止一个，起点多就要多次调用函数，因此超时！

倒不如反向建图，变成一个起点，多个终点！

另外注意：公车车是单向的！

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAX 0x3f3f3f3f
using namespace std;
int map[1010][1010],d[1010],n,m,s,t;
void dijkstra(int x)
{
    int i,j,min,mark,used[1010];
    for(i=1;i<=n;i++)
    {
        used[i]=0;
        d[i]=map[x][i];
    }
    d[x]=0;
    used[x]=1;
    for(i=1;i<=n;i++)
    {
        min=MAX;
        mark=-1;
        for(j=1;j<=n;j++)
        {
            if(!used[j]&&d[j]<min)
            {
                min=d[j];
                mark=j;
            }
        }
        if(mark==-1)
        break;
        used[mark]=1;
        for(j=1;j<=n;j++)
        {
            if(!used[j]&&d[j]>d[mark]+map[mark][j])
            d[j]=d[mark]+map[mark][j];
        }
    }

}
int main()
{
    int a,b,c,i,j;
    while(scanf("%d%d%d",&n,&m,&s)!=EOF)
    {
        memset(map,MAX,sizeof(map));
        for(i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[b][a]>c)
            map[b][a]=c;
        }
        dijkstra(s);//调用一次
        int start;
        int mi=MAX;
        scanf("%d",&t);
        for(i=0;i<t;i++)//找出最小的花费
        {
            scanf("%d",&start);
            mi=mi<d[start]?mi:d[start];
        }
        if(mi==MAX)
        printf("-1\n");
        else
        printf("%d\n",mi);
    }
return 0;
}