Til the Cows Come Home poj 2387 c++

水深无声 2022-08-11 00:42 3阅读 0赞

Til the Cows Come Home

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 29656</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 9970</td> 
  </tr> 
 </tbody> 
</table>

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.  
  
Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.  
  
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

\* Line 1: Two integers: T and N  
  
\* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

\* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
    1 2 20
    2 3 30
    3 4 20
    4 5 20
    1 5 100

Sample Output

经典的dijskra最短路径题：

迪杰斯特拉 基本思想是  ：  
 我们在N个点中\{1,2…..N\}，找从点1出发到各个点的最短路径。  
V=\{2，….,N\};S=\{1\};//初始状态  
我们依次从集合V中找到离S最近的点，然后把点Vi加入S中，然后用点Vi去刷新集合V中点到起点的距离(也即V1经过Vi到达集合V中的点),依次循环直至所有点都加入S。

AC代码：

#include<cstdio>
    #include<cstring>
    #define MIN 1000000000
    
    int point[1010][2020],dist[1100],flag[1010];
    
    void dijskra(int n )
    {
         int i,j,k,v;
         for(i=2;i<=n;i++)
             dist[i]=point[1][i];
         dist[1]=0;    
         flag[1]=1;
         for(i=2;i<=n;i++)
         {
             int min=MIN ;  
             for(j=1;j<=n;j++)
                 if(!flag[j]&&dist[j]<min)
                 {
                     v=j;
                     min=dist[j];
                 }
             flag[v]=1;
             for(k=1;k<=n;k++)
                 if(!flag[k]&&(min+point[v][k]<dist[k]))  
                      dist[k]=min+point[v][k]  ;
         }
    }
    int main()
    {
        int i,j,n,m,t,a,b,c;
        while(scanf("%d%d",&t,&n)!=EOF)
        {
           
             memset(flag,0,sizeof(flag));//标记进入S的点； 
             for(i=0;i<=n;i++)
             {
                 dist[i]=MIN;//最小距离；//权值； 
                 for(j=0;j<=n;j++)
                 {
                     if(i==j)
                          point[i][j]=0;
                     else     
                          point[i][j]=MIN;//直接路径；
                 }
             }
             for(i=1;i<=t;i++)
             {
               scanf("%d%d%d",&a,&b,&c);
               if(point[a][b]>c)
               point[a][b]=point[b][a]=c;
             }
             dijskra(n);
             printf("%d\n",dist[n]);
        }
    }