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. 在Pandas中直接加载MongoDB的数据

    在使用Pandas进行数据处理的时候,我们通常从CSV或EXCEL中导入数据,但有的时候数据都存在数据库内,我们并没有现成的数据文件,这时候可以通过Pymongo这个库,从mongoDB中读取数据,然 ...

  2. BZOJ1021 [SHOI2008]循环的债务

    Description Alice.Bob和Cynthia总是为他们之间混乱的债务而烦恼,终于有一天,他们决定坐下来一起解决这个问题. 不过,鉴别钞票的真伪是一件很麻烦的事情,于是他们决定要在清还债务 ...

  3. react 使用 ref 报错 ,[eslint] Using string literals in ref attributes is deprecated. (react/no-string-refs)

    react 项目中给指定元素加事件,使用到 react 的 ref 属性,Eslink 报错 [eslint] Using string literals in ref attributes is d ...

  4. jQuery中判断input的disabled属性

    <input type="text" id="ipt1" disabled> <input type="text" id= ...

  5. bootstrap学习笔记细化(表单)

    主要属性: class="form-inline" 水平排列 class="form-group" 组键 form-control 圆角方框发光 input-l ...

  6. HDFS原理解析

    一.HDFS简介 HDFS为了做到可靠性(reliability)创建了多分数据块(data blocks)的复制(replicas),并将它们放置在服务器群的计算节点中(computer nodes ...

  7. Linux 中常用命令

    命令基本格式: 命令提示符:[root@localhost ~]#      root 代表当前的登录用户(linux当中管理员账号是root)      @ 无实际意义      localhost ...

  8. xfs参数简介

    age_buffer_centisecs age_buffer_centisecs:(Min: 100  Default: 1500  Max: 720000) 多长时间设置为脏数据 xfsbufd_ ...

  9. 安卓app开发-02-安卓app快速开发

    安卓app开发-02-安卓app快速开发 上一篇介绍了安卓 app 开发的工具和环境配置,本篇不涉及编程技术,适合小团队快速高效开发 APP制作流程 当有一个APP创意,该如何实现呢?是花数十万找AP ...

  10. JS Error 内置异常类型 处理异常 Throw语句

    Exceptional Exception Handling in JavaScript       MDN资料 Anything that can go wrong, will go wrong. ...