HDU 2680 最短路径

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5255    Accepted Submission(s): 1673

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 51 2 21 5 31 3 42 4 72 5 62 3 53 5 14 5 122 34 3 41 2 31 3 42 3 211

 

Sample Output

1-1
模板..
#include <iostream>using namespace std;const int maxn=5555;  //注意数组下标const int inf=9999999;int map[maxn][maxn];int dis[maxn];//存放当前源点V到其他点i的最短距离 bool vis[maxn];int n,m;//顶点数 int start,endd;//起点终点void Dijkstra(int x)//x为源点 {     int i,j,k,Min;    memset(vis,0,sizeof(vis));     for(i=0;i<n;i++)                //初始化dist         dis[i]=map[x][i];     vis[x]=1;     k=0;             //K初始化关键！    for(i=1;i<n;i++) //数组小标从0开始的存图！    {              Min=inf;        for(j=0;j<n;j++)        {                if(!vis[j]&&dis[j]<Min)            {                      k=j;                Min=dis[j];            }        }        if(Min==inf)            break;        vis[k]=1;        for(j=0;j<n;j++)        {            if(!vis[j]&&dis[j]>dis[k]+map[k][j])                dis[j]=dis[k]+map[k][j];        }    }}int main(){    int i,j,t,x;    int a,b,c;    while(~scanf("%d%d%d",&n,&m,&endd))    {        for(i=0;i<n;i++)            for(j=0;j<n;j++)                map[i][j]=inf;        for(i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            if(map[b-1][a-1]>c)                map[b-1][a-1]=c;        }        int minn=inf;        Dijkstra(endd-1);        scanf("%d",&t);        for(i=0;i<t;i++)        {            scanf("%d",&x);            if(dis[x-1]<minn)                minn=dis[x-1];        }        if(minn==inf)            printf("-1\n");        else            printf("%d\n",minn);    }    return 0;}