hdu acm 1115 Lifting the Stone（多边形重心）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6851    Accepted Submission(s): 2860

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

245 00 5-5 00 -541 111 111 111 11

 

Sample Output

0.00 0.006.00 6.00

 

解题关键：

1.   三角形重心：x=(x0+x1+x3)/3，y=(y0+y1+y2)/3; 
2.     已知一多边形没有边相交，质量分布均匀。顺序给出多边形的顶点坐标，求其重心。 
3.     将n+2多变形分成n个三角形，总面积为S，分重心为(xi,yi),分面积为si，则重心为X=(∑si*xi)/S,Y=(∑si*yi)/S;这个公式也适合凹多边形。
   

#include<iostream>#include<stdio.h>#include<queue>#include<string.h>#include<algorithm>#define maxn 1000005#define MS(a,b) memset(a,b,sizeof(a))using namespace std;double cross(double x1,double y1,double x2,double y2){    return x1*y2-x2*y1;//叉积求三角形面积的2倍。}double x[maxn],y[maxn],area,sum,sx,sy,x1,y1;int main(){   int t,n,i;   cin>>t;   while(t--)   {       cin>>n;    for(i=0;i<n;i++)     scanf("%lf %lf",&x[i],&y[i]);      sx=sy=sum=0;     for(i=2;i<n;i++)     {      area=0.5*cross(x[i-1]-x[0],y[i-1]-y[0],x[i]-x[0],y[i]-y[0]);       sum=sum+area;       sx=sx+area*(x[i]+x[0]+x[i-1]);//本来尾巴后面应该除以3，放在最后答案里除以3.       sy=sy+area*(y[i]+y[0]+y[i-1]);//本来尾巴后面应该除以3，放在最后答案里除以3.     }     printf("%.2lf %.2lf\n",sx/(3.0*sum),sy/(3.0*sum));   }  return 0;}