ACM--多边形重心–-HDOJ 1115-- Lifting the Stone

HDOJ题目地址：传送门

Lifting the Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7041    Accepted Submission(s): 2965

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
   
   
   
   

计算多边形重心方法：
（1）划分多边形为三角形：
以多边形的一个顶点V为源点（V可取输入的第一个顶点），作连结V与所有非相邻顶点的线段，即将原N边形或分为（N-2）个三角形；
（2）求每个三角形的重心和面积：
设某个三角形的重心为G（cx，cy），顶点坐标分别为A1（x1，y1），A2（x2，y2），A3（x3，y3），则有cx = (x1 + x2 + x3)/3.同理求得cy。求面积的方法是s =  ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2，当A1,A2,A3顺时针排列时取-，否则取正（此定理不证）。事实上，在求每个三角形时不需要辨别正负，之后有方法抵消负号，见下述。
（3）求原多边形的重心：
公式：cx = (∑ cx[i]*s[i]) / ∑s[i];  cy = (∑ cy[i]*s[i] ) / ∑s[i];其中（cx[i], cy[i]）, s[i]分别是所划分的第i个三角形的重心坐标和面积。由题“ connect the points in the given order”知每个s[i]的正负号相同，故而∑ cx[i]*s[i]能与∑s[i]消号，所以根本不需要在第（2）步判断每个s[i]的正负。另外，在（2）中求每个重心坐标时要除以3，实际上不需要在求每个三角形坐标时都除以3，只需要求出∑ cx[i]*s[i]后一次性除以3即可。即是多边形重心坐标变为：cx = (∑ cx[i]*s[i]) / （3*∑s[i]）;  cy = (∑ cy[i]*s[i] ) / （3*∑s[i]）;


总结：
每个三角形重心：cx = x1 + x2 + x3；cy同理。
每个三角形面积：s =  ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2；
多边形重心：cx = (∑ cx[i]*s[i]) / （3*∑s[i]）;  cy = (∑ cy[i]*s[i] ) / （3*∑s[i]）;

#include <iostream>
#include <cstdio>
using namespace std;
int T, N, x[3], y[3];
double sumx, sumy, sumarea;
int main()  {
    scanf("%d", &T);
    while(T--){
        sumx = sumy = sumarea = 0.0;
        scanf("%d", &N);
        scanf("%d %d %d %d", x, y, x+1, y+1);
        N -= 2;
        while(N--){
            scanf("%d %d", x+2, y+2);
            //求新添加的三角形的面积
            double s = ((x[1] - x[0]) * (y[2] - y[0]) - (x[2] - x[0]) * (y[1] - y[0])) / 2.0;
            //求∑x[i]*s[i]和∑y[i]*s[i]
            sumx += ((x[0] + x[1] + x[2]) * s);
            sumy += ((y[0] + y[1] + y[2]) * s);
            //求总面积
            sumarea += s;

            x[1] = x[2];
            y[1] = y[2];
        }
        printf("%.2lf %.2lf\n", sumx / sumarea / 3, sumy / sumarea / 3);
    }
    return 0;
}

参考博客：http://blog.csdn.net/cs_zlg/article/details/7870799