hdu-2680 Choose the best route(最短路)

题目链接：

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 32768/32768 K (Java/Others)

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

 

题意：

 

给一个有向图,问从多个起点的任意一个出发到达终点的最短时间;

 

思路：

 

把0当做所有起点的起点,那么这些起点到0的距离都是0,这样可以用dijkstra算法跑一波得到答案,如果用Floyd算所有节点对应该会tle;

 

AC代码：

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const int N=1e5+6;
const int inf=0x3f3f3f3f;
int n,m,s;
int p[1002][1002],flag[1002],dis[1002];
void dijkstra()
{
    memset(flag,0,sizeof(flag));
    for(int i=0;i<=n;i++)
    {
        dis[i]=p[0][i];
    }
    flag[0]=1;
    int temp;
    for(int i = 0;i <= n;i++)
    {
        int mmin=inf;
        for(int j = 1;j<=n;j++)
        {
            if(!flag[j]&&dis[j]<mmin)
            {
                mmin=dis[j];
                temp=j;
            }
        }
        if(mmin == inf)break;
        flag[temp]=1;
        for(int j=1;j<=n;j++)
        {
            if(dis[j]>dis[temp]+p[temp][j])
                dis[j]=dis[temp]+p[temp][j];
        }
    }
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&s)!=EOF)
    {
        for(int i=0;i<=n;i++)
        {
            dis[i]=inf;
            for(int j=0;j<=n;j++)
            {
                if(i == j)p[i][j]=0;
                else p[i][j]=inf;
            }
        }
        int u,v,w;
        for(int i = 0;i < m;i ++)
        {
            scanf("%d%d%d",&u,&v,&w);
            p[u][v]=min(p[u][v],w);
        }
        int num,x;
        scanf("%d",&num);
        for(int i=0;i<num;i++)
        {
            scanf("%d",&x);
            p[0][x]=0;
        }
        dijkstra();
        if(dis[s] == inf)printf("-1
");
        else printf("%d
",dis[s]);
    }

    return 0;
}