hdu 2680 Choose the best route(最短路dijkstra(有重边))

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8477    Accepted Submission(s): 2780

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

题目分析：

注意有重边，所以一定要利用邻接矩阵存边，然后只存两点间最小的边，邻接链表不能保证这一特性，邻接链表排一遍序也可行，每次先跑同一种边中较小的

#include <algorithm>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#define MAX 1007

using namespace std;

int n,m,s,k;
int mp[MAX][MAX];
int dis[MAX];
bool used[MAX];

struct cmp
{
    bool operator ( ) ( const int&a , const int&b ) const 
    {
        return dis[a] > dis[b];
    }
};

void dijkstra ( )
{
    memset ( dis , 0x3f , sizeof ( dis ) );
    memset ( used , 0 , sizeof ( used ) );
    priority_queue<int , vector<int> , cmp > q;
    dis[s] = 0;
    q.push ( s );
    while ( !q.empty( ) )
    {
        int u = q.top();
        q.pop();
        used[u] = true;
        for ( int v = 1 ; v <= n ; v++ )
            if ( !used[v]&&dis[v] > dis[u] + mp[u][v] )
            {
                dis[v] = dis[u] + mp[u][v];
                q.push ( v );
            }
    }
}

int main ( )
{
    int u , v , w;
    while ( ~scanf ( "%d%d%d" , &n , &m , &s ) )
    {
        memset ( mp , 0x3f , sizeof ( mp ) );
        const int INF = mp[0][0];
        for ( int i = 0 ; i < m ; i++ )
        {
            scanf ( "%d%d%d" , &v , &u , &w );
            mp[u][v] = min ( mp[u][v] , w ); 
        }
        dijkstra();
        scanf ( "%d" , &k );
        int ans = INF;
        while ( k-- )
        {
            scanf ( "%d" , &v );
            ans = min ( ans , dis[v] );
        }
        if ( ans == INF ) puts ( "-1" );
        else printf ( "%d\n" , ans );
    }
}