(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/Others)

Total Submission(s): 168 Accepted Submission(s): 98

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

Central Europe 1999

 

Recommend

Eddy

 

题目分析：

求凸多边形的重心，简单题。

1、　累加和求重心
设平面上有N 个离散数据点( xi , yi ) ( i = 1, 2, ., n) , 其
多边形重心G( . x1, . y1) 为:

　　

　　这是求多边形最简单直观的方法。能够直接利用离散数
据点的x, y坐标就能求图形重心。

可是缺陷在于没有对离散
数据点所围图形做不论什么处理和分析,精度不够。

2、　算法一：在讲该算法时，先要明确以下几个定理。

定理1　已知三角形△A1A2A3的顶点坐标Ai ( xi , yi ) ( i =1, 2, 3) 。它的重心坐标为:

　　xg = (x1+x2+x3) / 3 ;                       yg = (y1+y2+y3) / 3 ;

定理2　已知三角形△A1A2A3的顶点坐标Ai ( xi , yi ) ( i =1, 2, 3) 。

该三角形的面积为:

　　S =  ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2 ;

　　△A1A2A3 边界构成逆时针回路时取+ , 顺时针时取 -。

　　另外在求解的过程中，不须要考虑点的输入顺序是顺时针还是逆时针，相除后就抵消了。

　　原理:将多边形划分成n个小区域, 每一个小区域面积为σi ,重心为Gi ( . xi , . yi ) ,利用求平面薄板重心公式把积分变
　　成累加和:

　　　　

　　　　　　　　　　　　　　　　　　

　　　　由前面所提出的原理和数学定理能够得出求离散数据点所围多边形的一般重心公式:以Ai ( xi , yi ) ( i = 1, 2, ., n) 为顶点的随意N边形A1A2 .An ,将它划　　　　分成N - 2个三角形(如图1) 。每一个三角形的重心为Gi ( . xi , . yi ) ,面积为σi。那么多边形的重心坐标G( .x2, .y2) 为:

　　

图1  多边形分解

代码例如以下：

#include <iostream>
#include <cstdio>

using namespace std;

const int maxn = 1000001;

struct PPoint {
	double x, y;
};

double Area(PPoint p0, PPoint p1, PPoint p2) {
	double area = 0;
	area = p0.x * p1.y + p1.x * p2.y + p2.x * p0.y - p1.x * p0.y - p2.x * p1.y
			- p0.x * p2.y;
	return area / 2;  // 另外在求解的过程中，不须要考虑点的输入顺序是顺时针还是逆时针。相除后就抵消了。

}

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);

		PPoint p0,p1,p2;

		scanf("%lf %lf",&p0.x,&p0.y);
		scanf("%lf %lf",&p1.x,&p1.y);

		double sum_area = 0;
		double sum_x = 0;
		double sum_y = 0;

		int i;
		for(i = 2 ; i < n ; ++i){
			scanf("%lf %lf",&p2.x,&p2.y);

			double area = Area(p0,p1,p2);
			sum_area += area;
			sum_x += (p0.x + p1.x+ p2.x)*area;
			sum_y += (p0.y + p1.y + p2.y)*area;
			p1 = p2;
		}

		printf("%.2lf %.2lf\n",sum_x/(sum_area*3),sum_y/(sum_area*3));
	}

	return 0;
}