Eddy's picture


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10218    Accepted Submission(s): 5162


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


Author

eddy

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如果能用一句话形容我现在的心情  我真想说  

一天一道题没做出来  只能虐虐水题  心情不爽 很想发泄

#include <stdio.h>
#include <math.h>
#define inf 0x3fffffff
struct node
{
  double x,y;
}p[105];
int n;
double line[105][105];
double getDist(node p1,node p2)
{
  return sqrt(pow(p1.y-p2.y,2)+pow(p1.x-p2.x,2));
}
void prim()
{
  double dist[105];
  for(int i=0;i<n;i++)
  dist[i]=inf;
  int now=0;
  double ans=0;
  for(int i=0;i<n-1;i++)
  {
    for(int j=0;j<n;j++)
    {
      if(dist[j]!=-1&&dist[j]>line[now][j])
      {
        dist[j]=line[now][j];
      }
    }
    
    dist[now]=-1;
    double min=inf;
    int x;
    for(int j=0;j<n;j++)
    {
      if(dist[j]!=-1&&min>dist[j])
      {
        min=dist[j];
        x=j;
      }
    }
    now=x;
    ans+=min;
  }
  printf("%.2lf\n",ans);
}
int main()
{
  while(~scanf("%d",&n))
  {
    for(int i=0;i<n;i++)
    {
      scanf("%lf %lf",&p[i].x,&p[i].y);
    }
    for(int i=0;i<n;i++)
    {
      line[i][i]=inf;
      for(int j=i+1;j<n;j++)
      {
        line[i][j]=line[j][i]=getDist(p[i],p[j]);
      }
    }
    prim();
  }
}