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Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 28   Accepted Submission(s) : 17

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Problem Description

Input

Output

Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

这道题也是很简单的最短路径问题

代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#define INF 1 << 30
double a[101] , b[101] , map[101][101] ;
double dis[101] ;
int used[101] ;
void prim(int n)
{
 int c = 0 ;
 int i = 0 , j = 0 ;
 double sum = 0 ;
 dis[1] = 0 ;
 for(i = 1 ; i <= n ; i++)
 {
  double min = INF ;
  c = 0 ;
  for(j = 1 ; j <= n ; j++)
  {
         if(!used[j] && dis[j] < min)
   {
    min = dis[j] ;
    c =j ;
   }
  }
  used[c] = 1 ;
  for(j  = 1 ; j <= n ; j++ )
  {
   if(!used[j] && dis[j] > map[c][j])
    dis[j] = map[c][j] ;
  }
 }
 
 for(i = 1 ; i <= n ; i++)
  sum += dis[i] ;
 printf("%.2lf\n",sum);
}
int main()
{
 int n = 0 ;
 while(~scanf("%d",&n))
 {
  memset(a , 0 , sizeof( a ) ) ;
  memset(b , 0 , sizeof( b ) ) ;
  int i = 0 , j = 0 ;
  for(i = 1 ; i <= n ; i++)
  {
   for(j = 1 ; j <= n ; j++)
       map[i][j] = INF ;
   dis[i] = INF ;
   used[i] = 0 ;
  }
  for(i = 1 ; i <= n ; i++)
  {
            scanf("%lf%lf" , &a[i] , &b[i] );
  }
  double m = 0 , x = 0;
  for(i = 1 ; i <= n ; i++ )
  {
            for(j = 1 ; j <= n ; j++)
   {
    x = (a[j]-a[i])*(a[j]-a[i])+(b[j]-b[i])*(b[j]-b[i]) ;
    m = sqrt( x ) ;
    map[i][j] = map[j][i] = m ;
   }
  }
  prim( n );
 }
 return 0 ;
}

杭电1162Eddy's picture

标签：杭电   1162   eddys picture   

原文地址：http://blog.csdn.net/bluedream1219/article/details/37989707