Eddy's picture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11841    Accepted Submission(s): 5922

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

Recommend

JGShining   |   We have carefully selected several similar problems for you:  1217 1875 1879 1863 1325

分析:

最小生成树问题(采用克鲁斯卡尔算法)

不过边还需要自己求(两点间的距离公式,点的组合)

code:

#include<bits/stdc++.h>
using namespace std;
#define max_v 105
struct edge
{
int x,y;
double w;
};
edge e[max_v*max_v];
int pa[max_v],rk[max_v];
double sum=0.0;
bool cmp(edge a,edge b)
{
return a.w<b.w;
}
void make_set(int x)
{
pa[x]=x;
rk[x]=;
}
int find_set(int x)
{
if(x!=pa[x])
pa[x]=find_set(pa[x]);
return pa[x];
}
void union_set(int x,int y,double w)
{
x=find_set(x);
y=find_set(y);
if(x==y)
return ;
if(rk[x]>rk[y])
pa[y]=x;
else
{
if(rk[x]==rk[y])
rk[y]++;
pa[x]=y;
}
sum+=w;
return ;
}
double f(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
int n;
while(~scanf("%d",&n))
{
sum=;
double a[max_v],b[max_v];
for(int i=; i<n; i++)
{
make_set(i);
scanf("%lf %lf",&a[i],&b[i]);
}
int k=;
for(int i=; i<n; i++)
{
for(int j=i+; j<n; j++)
{
e[k].x=i;
e[k].y=j;
e[k++].w=f(a[i],b[i],a[j],b[j]);
}
}
sort(e,e+k,cmp);
for(int i=; i<k; i++)
{
union_set(e[i].x,e[i].y,e[i].w);
}
printf("%0.2lf\n",sum);
}
return ;
}

HDU 1162Eddy's picture(MST问题)的更多相关文章

  1. 【HDU 1828】 Picture (矩阵周长并,线段树,扫描法)

    [题目] Picture Problem Description A number of rectangular posters, photographs and other pictures of ...

  2. hdu Eddy's picture (最小生成树)

    Eddy's picture Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Tota ...

  3. HDU 5627 Clarke and MST &意义下最大生成树 贪心

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5627 题意:Bestcoder的一道题,让你求&意义下的最大生成树. 解法: 贪心,我们从高位 ...

  4. hdu 1679 The Unique MST (克鲁斯卡尔)

    The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24152   Accepted: 8587 D ...

  5. 【49.23%】【hdu 1828】Picture

    Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s) ...

  6. hdu 1863 - 畅通工程(MST)

    畅通工程 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submis ...

  7. hdu 5627 Clarke and MST(最大 生成树)

    Problem Description Clarke is a patient with multiple personality disorder. One day he turned into a ...

  8. HDU - 4786 Fibonacci Tree (MST)

    题意:给一张由白边和黑边构成的无向图,求是否存在一个生成树,使白边的数量为一个斐波那契数. 分析:白边权值为1,黑边权值为0.求出该图的最小生成树和最大生成树,若这两个值之间存在斐波那契数,则可以,若 ...

  9. HDU 1828:Picture(扫描线+线段树 矩形周长并)

    题目链接 题意 给出n个矩形,求周长并. 思路 学了区间并,比较容易想到周长并. 我是对x方向和y方向分别做两次扫描线.应该记录一个pre变量,记录上一次扫描的时候的长度,对于每次遇到扫描线统计答案的 ...

随机推荐

  1. zoj 1760 查找

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=760 撸了个二分查找 #include<iostream> #inc ...

  2. 1.springIOC初识

    IOC,控制反转,从最浅显的角度来讲就是通过Spring容器来负责创建对象 大体的实现结构 1.首先有一个我们需要运行的类 2.在spring专属的xml配置文件中配置该类 3.启动容器 4.从该容器 ...

  3. Java 基础:数组

    一.数组声明: int[] x; int x[]; 在Java中一般使用前者,机把int[]看做一个类型,C++中只能后者 二.数组初始化: 直接提供值: int[] x = {1, 3, 4}; i ...

  4. Bzoj2395: [Balkan 2011]Timeismoney(最小乘积生成树)

    问题描述 每条边两个权值 \(x,y\),求一棵 \((\sum x) \times (\sum y)\) 最小的生成树 Sol 把每一棵生成树的权值 \(\sum x\) 和 \(\sum y\) ...

  5. MySQL数据库更改默认引擎为Innodb【配置】

    InnoDB和MyISAM是许多人在使用MySQL时最常用的两个表类型,这两个表类型各有优劣,视具体应用而定. 基本的差别为:MyISAM类型不支持事务处理等高级处理,而InnoDB类型支持.MyIS ...

  6. C++学习笔记(7)----类的数组中构造函数和析构函数的调用顺序

    C++类的数组中构造函数和析构函数的调用顺序(2) 对于如下的代码: #include<iostream> using namespace std; class CBase { priva ...

  7. MySQL 备份恢复(导入导出)单个 innodb表

    MySQL 备份恢复单个innodb表呢,对于这种恢复我们我们很多朋友都不怎么了解了,下面一起来看一篇关于MySQL 备份恢复单个innodb表的教程 在实际环境中,时不时需要备份恢复单个或多个表(注 ...

  8. EF Migrations

    Enable-Migrations -EnableAutomaticMigrations dbcontent Add-Migration XXXXX Update-Database -Verbose ...

  9. Software Testing Techniques Homework 3

    1. a.This is the chart b. initial numPrimes = 4, t1 would over the loop. c. t = ( n = 1) d. node cov ...

  10. SpringMVC笔记——SSM框架搭建简单实例

    落叶枫桥 博客园 首页 新随笔 联系 订阅 管理 SpringMVC笔记——SSM框架搭建简单实例 简介 Spring+SpringMVC+MyBatis框架(SSM)是比较热门的中小型企业级项目开发 ...