zoukankan      html  css  js  c++  java
  • hdoj 1162 Eddy's picture

    并查集+最小生成树

    Eddy's picture

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7669    Accepted Submission(s): 3882


    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
     
    kruskal算法
    #include<stdio.h>
    #include<algorithm>
    #include<math.h>
    using namespace std;
    int set[110];
    struct record//注意此题要用实型
    {
    	double a;     
    	double b;
    	double ju;//两点之间距离 
    }s[10000];
    int find(int fa)
    {
    	int ch=fa;
    	int t;
    	while(fa!=set[fa])
    	fa=set[fa];
    	while(ch!=fa)
    	{
    		t=set[fa];
    		set[ch]=fa;
    		ch=t;
    	}
    	return fa;
    }
    void mix(int x,int y)
    {
    	int fx,fy;
    	fx=find(x);
    	fy=find(y);
    	if(fx!=fy)
    	set[fx]=fy;
    }
    double dis(double x1,double y1,double x2,double y2)
    {
    	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); //计算两点之间距离 
    }
    bool cmp(record c,record d)
    {
    	return c.ju<d.ju;   
    }
    int main()
    {
    	int m,j,i,point;
    	double sum;
    	double a[110];//这两个数组用来储存点的坐标,a数组储存横标 
    	double b[110];//b数组储存纵标 
    	while(scanf("%d",&point)!=EOF)
    	{
    		for(i=0;i<=point;i++)
    		set[i]=i;
    		for(i=1;i<=point;i++)
    		{
    			scanf("%lf%lf",&a[i],&b[i]);
    		}
    		m=0;
    		for(i=1;i<=point-1;i++)    //此循环求任意两点之间距离 
    		{                          //并记录下此两点位置 
    			for(j=i+1;j<=point;j++)
    			{
    				s[m].ju=dis(a[i],b[i],a[j],b[j]);
    				s[m].a=i;
    				s[m].b=j;
    				m++;				
    			}
    		}
    	    sort(s,s+m,cmp);
    		sum=0;
    		for(i=0;i<m;i++)
    		{
    			if(find(s[i].a)!=find(s[i].b))
    			{
    				mix(s[i].a,s[i].b);
    				sum+=s[i].ju;
    			}
    		}
    		printf("%.2lf
    ",sum);
    	}
        return 0;	
    }
    

      

     
  • 相关阅读:
    echarts曲线图
    echarts画柱状图
    echarts画环形图
    ppt素材网
    黄元御的桔梗元参汤治疗过敏性鼻炎
    vue双向数据绑定对于数组和新增对象属性不能监听的解决办法
    谈谈vue双向数据绑定问题
    一个不错的中医博客
    javascript 中 keyup、keypress和keydown事件
    浏览器渲染页面的过程
  • 原文地址:https://www.cnblogs.com/tonghao/p/4470938.html
Copyright © 2011-2022 走看看