HDOJ Eddy's picture-优快云博客

Eddy热爱画画并希望用一个编程问题来改变朋友们对他画技的看法。问题要求计算平面上若干点通过直线相连所需的最短墨水总量，采用Kruskal算法解决最小生成树问题。

Eddy's picture

http://acm.hdu.edu.cn/diy/contest_showproblem.php?pid=1009&cid=12467&hide=0

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

Total Submission(s) : 26 Accepted Submission(s) : 8

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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

Author

eddy

代码如下：

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
struct{
        double x,y;
        int parent;
}point[105];
struct Edge{
       int from,to;
       double dis;
}edge[5100];
void init(int n){
     for(int i=1;i<=n;i++)
     point[i].parent=i;
}
int find(int x){
    while(x!=point[x].parent)
    x=point[x].parent;
    return x;
}
void merge(int x,int y){
     x=find(x);
     y=find(y);
     if(x!=y)
     point[x].parent=y;
}
bool compare(Edge a,Edge b){
     return a.dis<b.dis;
}
int main(){
    int n;
    double sum;
    while(scanf("%d",&n)!=EOF){
          float a,b;
           int p=1;
           init(n);
          for(int i=1;i<=n;i++)
            cin>>point[i].x>>point[i].y;
          for(int i=1;i<=n;i++){
             for(int j=i+1;j<=n;j++){
                    edge[p].from=i;
                    edge[p].to=j;
        edge[p].dis=sqrt( pow((point[i].x-point[j].x),2)+pow((point[i].y - point[j].y),2));
                    p++;
                          }
                  }

           sort(edge+1,edge+p,compare);
           int m=p-1,x,y;
           sum=0.0;
           for(int i=1;i<=m&&n>1;i++){
                    x=find(edge[i].from);
                    y=find(edge[i].to);
                   if(x!=y){
                       merge(x,y);
                       n--;
                       sum+=edge[i].dis;
                            }
                   }
           printf("%.2lf\n",sum);

     }
return 0;
}