Eddy's picture


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9101    Accepted Submission(s): 4607


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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题意：给你n个点，用最短的距离将这n个点连接起来。




题解：其实就是求最短路，点连通最短路。（实际上是求最小权的生成树，即最小生成树）


          可以上 prim，并查集，dijkstra，kruskal 等算法。。。。




AC代码：


​ 

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;

#define INF 0x3ffffff
struct point
{
  double x,y;
};
point p[105];
int n;
double ans;
double map[105][105];

void prim()
{
  double dis[105];
  double min;
  int k;
  dis[1]=0;
  for(int j=2;j<=n;j++)
  {
    dis[j]=map[1][j]; //其他点的距离 
  }
  for(int i=2;i<=n;i++) //找其他n条边 
  {
    min=INF;
    for(int j=1;j<=n;j++) 
    {
      if(dis[j]<min && dis[j]) //有可能重复 ,标志 
      {
        min=dis[j];
        k=j;
      }          
    }
      ans+=min; 
    dis[k]=0;
    for(int j=1;j<=n;j++)
    {
      if(map[k][j]< dis[j] && dis[j]) 
        dis[j]=map[k][j];

     } 
  }
}
int main()
{

  while(~scanf("%d",&n))
  {
    ans=0;
    for(int i=1;i<=n;i++)
    {
      scanf("%lf %lf",&p[i].x,&p[i].y);
    }
    for(int i=1;i<=n;i++)
    {
      for(int j=1;j<=n;j++)
      {
        if(i==j)map[i][j]=INF;
        else map[i][j]=sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));

      }
    }
    prim();
    printf("%.2f\n",ans);
  }
  return 0;
}