hdu 1162Eddy's picture Prime算法求最小生成树_picture问题及算法设计-优快云博客

本文介绍了一个有趣的编程问题：如何利用最小生成树算法帮助Eddy连接画纸上的所有点，并找到最短的连线总长度。通过解析问题背景及提供具体的C语言实现代码，展示了Prime算法的应用。

Eddy's picture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2153 Accepted Submission(s): 1023

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
典型的Prime算法求最小生成树，只要将数据处理一下就好，注意数据类型是double!!!!
#include<stdio.h>
#include<string.h>
#include<math.h>
#define maxdis 999999
double map[105][105];
double point[105][2];
int s[105];
double lowest[105],closest[105];
double dis(int i,int j)
{
    double d=sqrt((point[i][1]-point[j][1])*(point[i][1]-point[j][1])+
                  (point[i][0]-point[j][0])*(point[i][0]-point[j][0]));
    return d;
}
void calcu(int n)
{
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
      {
          map[i][j]=map[j][i]=dis(i,j);
      }
}
void process(int n)
{
    memset(s,0,sizeof(s));
    for(int i=1;i<=n;i++)
    { lowest[i]=map[i][1];
      closest[i]=1;
    }
    s[1]=1;
    double sum=0;
    for(int i=1;i<n;i++)
    {
        int u;
        int min=maxdis;
        for(int j=1;j<=n;j++)
        if(!s[j]&&lowest[j]<min){min=lowest[j];u=j;}
        s[u]=1;
        sum+=lowest[u];
        for(int k=1;k<=n;k++)
        if(!s[k]&&lowest[k]>map[k][u]) {lowest[k]=map[k][u];closest[k]=u;}
    }
    printf("%.2lf/n",sum);
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
        scanf("%lf%lf",&point[i][0],&point[i][1]);
        calcu(n);
        process(n);
    }
    return 0;
}