杭电1162Eddy's picture

Eddy's picture

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

Eddy begins to like  painting pictures recently ,he is sure of himself to become a painter.Every day Eddy draws pictures in his small room,  and he usually  puts out his newest pictures to let his friends appreciate. but the result it can be
imagined, the friends are not interested in his picture.Eddy feels very puzzled,in order to change all friends 's view to his technical of painting  pictures ,so Eddy creates a problem for the his friends of you.

Problem descriptions as follows: Given you some coordinates pionts on a drawing paper, every point links with the ink with the straight line, causes all points finally  to link in the same place. How many distants does your duty discover the shortest length
which the ink draws?

Input

The first line contains 0 < n <= 100, the number of point. For each point, a line follows; each following line contains two real numbers indicating the (x,y) coordinates of the point.




Input contains multiple test cases. Process to the end of file.

Output

Your program prints a single real number to two decimal places: the minimum total length of ink lines that can connect all the points.

Sample Input

3
1.0 1.0
2.0 2.0
2.0 4.0

Sample Output

3.41

这道题也是非常easy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 ;

}