hdu 2680 Choose the best route【dijstra+反向建图】

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15494    Accepted Submission(s): 5027

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= Then follow m lines ,each line contains three integers p , q , t (0 Then a line with an integer w(0

 

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

题意：给出顶点数和路径总数及要到达的终点，再给出可以到达终点的多个起点，输出这些起点中到终点的最短路径。

思路：如果找每个起点到终点的距离，最后比较出最小值，会tle（亲测有效），正解思路是反向建图，把题目所给的终点当作起点，多个起点当作终点，用一次dijstra求起点到每个顶点的最短路径，最后遍历查找起点到每个终点的路径的最短路。

看了别人的题解才知道，还有一个地方比较坑，同一路径，取权值最小的边，不过平时自己就习惯了判断自重边，反正也不影响结果，想不到这次避开了这个坑点。

#include
#include
#include<string.h>
#include
using namespace std;
#define inf 0x3f3f3f3f
#define N 1100
int w[N][N],dis[N],book[N],num[N];
int main()
{
     int n,m,s,t1,t2,t3,ans,min1,u,x,i,j;
     while(scanf("%d%d%d",&n,&m,&s)!=EOF)
     {
         for(i = 0; i <= n; i ++)
             for(j = 0; j <= n; j ++)
                     w[i][j] = inf;
        for(i = 1; i <= m; i ++)
        {
            scanf("%d%d%d",&t1,&t2,&t3);
            if(t3 < w[t2][t1])//判断自重边 
                w[t2][t1] = t3;//反向建图 
        }
         scanf("%d",&x);
         for(i = 1; i <= x; i ++)
             scanf("%d",&num[i]);
         memset(book,0,sizeof(book));
         for(i = 1; i <= n; i ++)
             dis[i] = w[s][i];
         book[s] = 1;
         for(i = 1; i < n; i ++)
         {
             min1 = inf;
             for(j = 1; j <= n; j ++)
                 if(!book[j]&&dis[j]<min1)
                 {
                     min1 = dis[j];
                     u = j;
                 }
             book[u] = 1;
             for(j = 1; j <= n; j ++)
                 if(!book[j]&&dis[j] > dis[u]+w[u][j])
                     dis[j] = dis[u]+w[u][j];
         }
         ans = inf;
         for(i = 1; i <= x; i ++)
            ans = min(ans,dis[num[i]]);//取反转图后起点到多个终点的最小权值 
        if(ans == inf)
             printf("-1\n");
         else
             printf("%d\n",ans);
 }
 return 0;
}