Lifting the Stone(多边形重心)

最新推荐文章于 2020-12-12 09:29:48 发布

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最新推荐文章于 2020-12-12 09:29:48 发布

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原文链接：http://www.cnblogs.com/haoabcd2010/p/5990927.html

本文介绍了一种通过将多边形分解为多个三角形来计算其重心的方法，并提供了一个具体的C++实现示例。

Lifting the Stone

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

Submit Practice HDU 1115

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

Sample Output

0.00 0.00
6.00 6.00


//求多边形的重心
第一行是案例数，然后是点的个数，然后是每个点的坐标
重量均匀分布的三角形，重心  X = (x1 + x2 + x3)/3 , Y = ( y1 + y2 + y3 )/3
质量集中在顶点上的多边形，n 个顶点坐标为(xi,yi)，质量为mi，则重心 
X = ∑( xi×mi ) / ∑mi 
Y = ∑( yi×mi ) / ∑mi
思路 : 将这个多边形转换成多个三角形，然后求出各个重心，将这些重心连起来形成个新多边形，求出重心
所以套公式就行了

 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 
 5 struct Node
 6 {
 7     double x,y;
 8 }node[100];
 9 
10 int main()
11 {
12     int n;
13     cin>>n;
14     while (n--)
15     {
16         int dian;
17         cin>>dian;
18         double x1,x2,y1,y2;
19         cin>>x1>>y1>>x2>>y2;
20 
21         int i;
22         double x,y;
23         double sumarea=0.0,sumx=0.0,sumy=0.0;
24         for (i=2;i<dian;i++)
25         {
26             cin>>x>>y;
27             double s=( (x2-x1) * (y-y1) - (x-x1) * (y2-y1) ) / 2;
28             // s= ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2
29             sumarea+=s;
30             sumx+=s*(x1+x2+x)/3;
31             sumy+=s*(y1+y2+y)/3;
32             x2=x;
33             y2=y;
34         }
35         printf("%.2lf %.2lf\n",sumx/sumarea,sumy/sumarea);
36     }
37     return 0;
38 }