杭电2680--Choose the best route…

Problem Description

One day , Kiki wants to visit one of her friends. As she isliable to carsickness , she wants to arrive at her friend’s home assoon 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 findout the least time Kiki needs to spend. To make it easy, if thecity have n bus stations ,the stations will been expressed as aninteger 1,2,3…n.

Input

There are several test cases.
Each case begins with three integers n, m ands,(n<1000,m<20000,1=<s<=n)n stands for the number of bus stations in this city and m standsfor the number of directed waysbetween bus stations .(Maybe there are several ways between two busstations .) s stands for the bus station that near Kiki’s friend’shome.
Then follow m lines ,each line contains three integers p , q , t(0<t<=1000). means from station p tostation 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. Thenfollows w integers stands for these stations.

Output

The output contains one line for each data set : the leasttime 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

 

注意题目中的红字，就是这个细节没注意到，Wa了n次！

# include<stdio.h>

#define MAX 100000000

int map[1008][1008],road[1008],time[1008];

int main()

{

      int m,i,j,x,y,z,min,d,n;

      while(scanf("%d%d%d",&n,&m,&d)!=EOF)

      {

             for(i=0;i<=n;i++)

                    for(j=0;j<=n;j++)

                             map[i][j]=MAX;

                    for(i=1;i<=m;i++)

                    {

                           scanf("%d%d%d",&x,&y,&z);

                           if(map[x][y]>z)

                                  map[x][y]=z;

                    }

                    scanf("%d",&y);

                    for(i=1;i<=y;i++)

                    {

                           scanf("%d",&x);

                           map[0][x]=0;

                           

                    }

                    for(i=0;i<=n;i++)//从哪点开始，就把该点到0点的距离置为0，用来统一起点，当终点不统一时也可用此方法！！

                    {

                           road[i]=0;

                           time[i]=map[0][i];

                    }

                    road[0]=1;

                    time[0]=0;

                    for(i=0;i<=n;i++)

                    {

                           min=MAX;

                           for(j=0;j<=n;j++)

                                  if(road[j]==0&&min>time[j])

                                  {

                                         min=time[j];

                                         x=j;

                                  }

                                  if(min==MAX)

                                         break;

                                  road[x]=1;

                                  for(j=0;j<=n;j++)

                                         if(road[j]==0&&time[j]>time[x]+map[x][j])

                                                time[j]=time[x]+map[x][j];

                    }

                    if(time[d]<MAX)

                           printf("%d\n",time[d]);

                    else

                           printf("-1\n");

      }

      return 0;

}