杭电OJ2680（dijkstra）choose the best route

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11170    Accepted Submission(s): 3627

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

这里首先提示大家：数组不要开的不恰当，我是TLE了几发，感觉心累。。。数组开小了不给RE给TLE。。。。真是够了。。。。

题目大意：（重点是单向图）

这里有n个车站，有m条路径，s是希望到达的终点（朋友的家），题干中一句：directed ways表示的是单向图，只能由车站a到车站b而不能从车站b开到车站a.（虽然题意是这样的，但是解题思路却是要改变这个定论的）.

这里我直接想到的方法还是简洁无脑的floyd做法，但是想了想1000个点绝壁不能floyd啊，但是可以尝试优化一下啊。然后小小的优化了一下，还是超时了，然后就默默的滚去找dijkstra的模板了。。。我作为一个智商比较低的孩纸，还是天真的尝试了一下用起点车站每一次都dijkstra，然后找到最小值。结果还是依旧的口怕。。。。。。。。。。。。。

这个时候就要想了，我有多起点是动态给出的，但是终点却是定点，机智的我这个时候用终点去找起点的最近路，秒掉了问题：

#include<stdlib.h>
#include<string.h>
#include<stdio.h>
using namespace std;
#define MAX(a,b) (a > b ? a : b)
#define MIN(a,b) (a < b ? a : b)
#define MAXN  10000001
#define INF 1000000007
int w[1010][1010];
int d[1010];
int vis[1010];
int qidian[1010];
int n,m,zongdian;
void dijkstra(int s)//dijkstra模板模板模板～～～～～～～～～～～。
{
    d[s]=0;
    for(int k=1;k<=n;k++)
    {
        int m=INF;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i] && d[i]<=m)
                m=d[s=i];
        }
        vis[s] = 1;
        for(int i=1;i<=n;i++)
        {
            d[i]=MIN(d[i],d[s]+w[s][i]);
        }
    }
}
int main()
{
    while(~scanf("%d%d%d",&n,&m,&zongdian))
    {
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            d[i]=INF;
            for(int j=1;j<=n;j++)
            {
                w[i][j]=INF;
            }
        }
        for(int i=0;i<m;i++)
        {
            int x,y,dis;
            scanf("%d%d%d",&x,&y,&dis);
            if(w[y][x]>dis)//注意是单向图，这个时候我要用终点找起点，所以是yx
            w[y][x]=dis;
        }
        dijkstra(zongdian);//拿终点去找起点。
       /* for(int i=1;i<=n;i++)
        {
            printf("%d ",d[i]);
        }
        printf("\n");*///测试数据专用~~~~~~~~~~~~~
        int q;
        int output=INF;
        scanf("%d",&q);
        for(int i=0;i<q;i++)
        {
            scanf("%d",&qidian[i]);
            if(d[qidian[i]]<output)output=d[qidian[i]];
        }
        if(output==INF){printf("-1\n");continue;}
        printf("%d\n",output);
    }
}