Lifting the Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5203    Accepted Submission(s): 2155

Problem 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
 
Source
 
数学几何,求多边形重心,乘上面积增加精度。
 
 //15MS    376K    695 B    C++
#include<stdio.h>
#include<math.h>
#define N 1000005
struct node{
double x,y;
}p[N];
double getsum(node a,node b,node c)
{
return ((b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y))/2.0;
}
int main(void)
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=;i<n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
double ss=,sx=,sy=;
for(int i=;i<n-;i++){
double ts=getsum(p[],p[i+],p[i+]);
ss+=ts;
sx+=ts*(p[].x+p[i+].x+p[i+].x);
sy+=ts*(p[].y+p[i+].y+p[i+].y);
}
printf("%.2lf %.2lf\n",sx/ss/,sy/ss/);
}
return ;
}

hdu 1115 Lifting the Stone (数学几何)的更多相关文章

  1. 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 ...

  2. 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 ...

  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. hdu1115 Lifting the Stone(几何,求多边形重心模板题)

    转载请注明出处:http://blog.csdn.net/u012860063 题目链接:pid=1115">http://acm.hdu.edu.cn/showproblem.php ...

  5. poj 1115 Lifting the Stone 计算多边形的中心

    Lifting the Stone Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

  6. hdu 1577 WisKey的眼神 (数学几何)

    WisKey的眼神 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  7. 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 ...

  8. (hdu step 7.1.3)Lifting the Stone(求凸多边形的重心)

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

  9. Lifting the Stone(多边形重心)

    Lifting the Stone Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u S ...

随机推荐

  1. MySql慢查询日志——开启/查看/删除

    1,开启慢查询日志 修改mysql.ini文件,加入如下配置: [mysqld] log-slow-queries=H:\mysql_log\slow_query.log long-query-tim ...

  2. LeetCode: 54. Spiral Matrix(Medium)

    1. 原题链接 https://leetcode.com/problems/spiral-matrix/description/ 2. 题目要求 给定一个二维整型数组,返回其螺旋顺序列表,例如: 最后 ...

  3. MyBatis-自定义结果映射规则

    1.自定义结果集映射规则 ①查询 <!-- public Employee getEmpById(Integer id); --> <select id="getEmpBy ...

  4. Selenium(Python) ddt读取Excel文件数据驱动

    首先, 引入xlrd模块: ExcelDDT.py: import unittestfrom time import sleep from ddt import ddt, datafrom selen ...

  5. Kotlin Android Extensions: 与 findViewById 说再见 (KAD 04) -- 更新版

    作者:Antonio Leiva 时间:Aug 16, 2017 原文链接:https://antonioleiva.com/kotlin-android-extensions/ 在 Kotlin1. ...

  6. ortp代码简析

    ortp初始化 /** *    Initialize the oRTP library. You should call this function first before using *     ...

  7. Python数学运算入门把Python当作计算器

    让我们尝试一些简单的 Python 命令.启动解释器,等待界面中的提示符,>>> (这应该花不了多少时间). 3.1.1. 数字 解释器就像一个简单的计算器一样:你可以在里面输入一个 ...

  8. 【转】Buff机制及其实际运用

    转自 http://bbs.gameres.com/forum.php?mod=viewthread&tid=215027 首先我想说的是,这是一套机制,并不是单独的一个系统,所谓机制就是一种 ...

  9. Python对文本文件逐行扫描,将含有关键字的行存放到另一文件

    #逐行统计关键字行数,并将关键字所在行存放在新的文件中 keyword = "INFO" b = open("C:\\Users\\xxx\\Documents\\new ...

  10. Java内存管理特点

    Java内存管理特点     Java一个最大的优点就是取消了指针,由垃圾收集器来自动管理内存的回收.程序员不需要通过调用函数来释放内存. 1.Java的内存管理就是对象的分配和释放问题.     在 ...