Choose the best route

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 86   Accepted Submission(s) : 35

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

————————————————————————————————————

题意：给出n,m,s，代表有n个车站，m条路线，s为终点

然后m行代表路线和花费，注意是单向

最后给出k个数代表能选择的出发点

思路：添加一个起点和所有出发点连0费边，Dijkstra求最短路 



#include <cmath>
#include <queue>
#include <string>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
#define inf 0x3f3f3f3f

int dis[1005];
int vis[1005];
int mp[1005][1005];
int m,n;

void djstl(int ed)
{
    for(int i=1; i<=n; i++)
    {
        dis[i]=mp[0][i];
        vis[i]=0;
    }
    dis[0]=0;
    vis[0]=1;
    for(int i=1; i<n; i++)
    {
        int mn=inf,u=-1;
        for(int j=1; j<=n; j++)
        {
            if(dis[j]<mn&&vis[j]==0)
            {
                mn=dis[j];
                u=j;
            }
        }
        if(u!=-1)
        {
            dis[u]=mn;
            vis[u]=1;
            for(int j=1; j<=n; j++)
            {
                if(dis[u]+mp[u][j]<dis[j]&&vis[j]==0)
                    dis[j]=dis[u]+mp[u][j];
            }

        }
    }
    if(dis[ed]<inf)
        printf("%d\n",dis[ed]);
    else
        printf("-1\n");
}



int main()
{
    int u,v,c,ed,k,o;
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        memset(mp,inf,sizeof(mp));
        for(int i=0; i<m; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            if(mp[u][v]>c)
                mp[u][v]=c;
        }
        scanf("%d",&o);
         while(o--)
        {
            scanf("%d",&ed);
           mp[0][ed]=0;
        }
        djstl(k);
    }
}