Lifting the Stone

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Description

There are many secret openings in the floor which are covered by a big heavy stone. When the stone is lifted up, a special mechanism detects this and activates poisoned arrows that are shot near the opening. The only possibility is to lift the stone very slowly and carefully. The ACM team must connect a rope to the stone and then lift it using a pulley. Moreover, the stone must be lifted all at once; no side can rise before another. So it is very important to find the centre of gravity and connect the rope exactly to that point. The stone has a polygonal shape and its height is the same throughout the whole polygonal area. Your task is to find the centre of gravity for the given polygon. 
 

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer N (3 <= N <= 1000000) indicating the number of points that form the polygon. This is followed by N lines, each containing two integers Xi and Yi (|Xi|, |Yi| <= 20000). These numbers are the coordinates of the i-th point. When we connect the points in the given order, we get a polygon. You may assume that the edges never touch each other (except the neighboring ones) and that they never cross. The area of the polygon is never zero, i.e. it cannot collapse into a single line. 
 

Output

Print exactly one line for each test case. The line should contain exactly two numbers separated by one space. These numbers are the coordinates of the centre of gravity. Round the coordinates to the nearest number with exactly two digits after the decimal point (0.005 rounds up to 0.01). Note that the centre of gravity may be outside the polygon, if its shape is not convex. If there is such a case in the input data, print the centre anyway. 
 

Sample Input

2 4 5 0 0 5 -5 0 0 -5 4 1 1 11 1 11 11 1 11
 

Sample Output

0.00 0.00 6.00 6.00
 
 
就是简单求多边形的重心问题,数学问题。
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
double x0,y0,x1,y1,x2,y2;
double s=0.0,sx=0.0,sy=0.0,area;
scanf("%lf%lf%lf%lf",&x0,&y0,&x1,&y1);
for(int i=;i<n-;i++)
{
scanf("%lf%lf",&x2,&y2);
area=(x1-x0)*(y2-y0)-(x2-x0)*(y1-y0);
sx+=area*(x0+x1+x2);
sy+=area*(y0+y1+y2);
s+=area;
x1=x2;
y1=y2;
// cout<<"dd"<<area<<endl; }
//cout<<s<<endl;
printf("%.2lf %.2lf\n",sx/s/,sy/s/);
}
return ;
}
附加两篇大神的博客,第一个是详细解释了poj1185 的算法,清晰明了
第二篇说的是求多边形重心的其他情况。
http://www.cnblogs.com/jbelial/archive/2011/08/08/2131165.html
 
 
http://www.cnblogs.com/bo-tao/archive/2011/08/16/2141395.html

poj 1115 Lifting the Stone 计算多边形的中心的更多相关文章

  1. POJ 1385 Lifting the Stone (多边形的重心)

    Lifting the Stone 题目链接: http://acm.hust.edu.cn/vjudge/contest/130510#problem/G Description There are ...

  2. hdu 1115:Lifting the Stone(计算几何,求多边形重心。 过年好!)

    Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others ...

  3. hdu 1115 Lifting the Stone 多边形的重心

    Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others ...

  4. Lifting the Stone(hdu1115)多边形的重心

    Lifting the Stone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...

  5. hdu 1115 Lifting the Stone (数学几何)

    Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others ...

  6. hdu 1115 Lifting the Stone

    题目链接:hdu 1115 计算几何求多边形的重心,弄清算法后就是裸题了,这儿有篇博客写得很不错的: 计算几何-多边形的重心 代码如下: #include<cstdio> #include ...

  7. [POJ 1385] Lifting the Stone (计算几何)

    题目链接:http://poj.org/problem?id=1385 题目大意:给你一个多边形的点,求重心. 首先,三角形的重心: ( (x1+x2+x3)/3 , (y1+y2+y3)/3 ) 然 ...

  8. Hdoj 1115.Lifting the Stone 题解

    Problem Description There are many secret openings in the floor which are covered by a big heavy sto ...

  9. POJ 3907 Build Your Home | 计算多边形面积

    给个多边形 计算面积 输出要四舍五入 直接用向量叉乘就好 四舍五入可以+0.5向下取整 #include<cstdio> #include<algorithm> #includ ...

随机推荐

  1. 普通用户如何临时获取root权限

    转自:http://634871.blog.51cto.com/624871/1325907 在实际工作中,公司不会将root用户直接给员工使用,而是通过员工自己的账号临时获得系统的root权限. 1 ...

  2. ubuntu显示桌面的快捷键,以及修改方法

    在ubuntu下面,快速显示桌面,你可以这样做. 1,ctrl+alt+d (默认的) 2,alt+tab 可以切换到桌面 但是我想把它修改成和windows一样的,我该怎么做呢? 其实很简单. 系统 ...

  3. Arbitrage(bellman_ford)

    Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16652   Accepted: 7004 Descri ...

  4. Android Bitmap 全面解析(四)图片处理效果对比 ...

    对比对象: UIL Volley 官方教程中的方法(此系列教程一里介绍的,ImageLoader的处理方法和官方的差不多) -------------------------------------- ...

  5. ssh和mvc理论基础

    ssh中mvc到底指的什么 mvcsshhibernatespringstrutsioc在SSH整合的架构中,Spring充当了一个容器的作用,Spring使用IOC和AOP技术接管了Hibernat ...

  6. Linux常用热键(持续更新)

    (这些文章都是从我的个人主页上粘贴过来的,大家也可以访问我的主页 www.iwangzheng.com) --圣诞节怎么过, --略过. 今天装ubuntu的时候把windows覆盖了, 凌乱,TX童 ...

  7. HDU 1708 简单dp问题 Fibonacci String

    Fibonacci String Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  8. Python配合BeautifulSoup读取网络图片并保存在本地

    本例为Python配合BeautifulSoup读取网络图片,并保存在本地. BeautifulSoup可代替正则表达式,更好地解析Html文本,获取其中的指定内容,如Tag.Property等 # ...

  9. php数组转换js数组操作及json_encode应用

    对于php,个人感觉能够熟练操作数组和字符串,基本上已经是入门了,php本身有很多操作数组和字符串的函数,今天在做一个功能时,需要用Js动态的创建门店信息,这些信息是要从后台添加的,想来想去,通过ph ...

  10. 【转】Velocity 语法

    一.基本语法 1."#"用来标识Velocity的脚本语句,包括#set.#if .#else.#end.#foreach.#end.#iinclude.#parse.#macro ...