Lifting the Stone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5440    Accepted Submission(s): 2278

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

这道题是求多边形重心的题目，可以做个模板。

该题目中的点顺序是按逆时针顺序给的。

求多边形重心分两种情况：

①质量集中在顶点上。n个顶点坐标为(xi,yi)，质量为mi，则重心

X = ∑( xi×mi ) / ∑mi        Y = ∑( yi×mi ) / ∑mi

特殊地，若每个点的质量相同，则    X = ∑xi / n        Y = ∑yi / n

②质量分布均匀。这个题就是这一类型，算法和上面的不同。

特殊地，质量均匀的三角形重心：

X = ( x0 + x1 + x2 ) / 3        Y = ( y0 + y1 + y2 ) / 3

题解参考自：http://blog.csdn.net/lttree/article/details/24720007

所以，这道题解法：

1.以一个点为顶点，做多个三角形，然后求每个三角形的重心与质量。

2.然后用每个三角形的重心为顶点再构成一个多边形。

3.这个多边形就属于上述的第一种情况了，质量在顶点上的多边形面积。

4.套公式就OK拉~

PS：注意几点：

由于三角形质量与面积成正比（WHY？因为质量分布均匀~。~），所以质量可用面积来代替。

再者用叉积求三角形面积要算正负情况（正负号要保留），

此举是为了防止多边形为凹多边形的情况，三角形面积会在多边形外。

OK~翠花，上模板~，此模板也会更新在我整理的     计算几何模板          中哟~~~

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#define x first
#define y second
using namespace std;

typedef long long LL;
typedef pair<double,double> Point;
const int N = 10010;
int n;
vector<Point>P;

Point cal()
{
    Point p ,s ;
    double tp , area = 0 , tpx = 0 ,tpy = 0;
    p.x = P[0].x ,p.y = P[0].y;
    for( int i = 1 ; i <= n  ;++i )
    {
        if(i == n ) s.x = P[0].x , s.y = P[0].y;
        else s.x = P[i].x , s.y = P[i].y;
        tp = ( p.x * s.y - p.y *s.x );
        area += tp / 2.0 ;
        tpx += ( p.x + s.x )* tp ;
        tpy += ( p.y + s.y )* tp ;
        p.x =s.x ; p.y = s.y;
    }
    s.x = tpx / (6*area);
    s.y = tpy / (6*area);
    return s;
}
void run()
{
    double xx,yy;
    P.clear();
    scanf("%d",&n);
    for(int i = 0 ; i< n ;++i){
        scanf("%lf%lf",&xx,&yy);
        P.push_back(Point(xx,yy));
    }
    Point res = cal();
    printf("%.2lf %.2lf\n",res.x,res.y);
}
int main()
{
    #ifdef LOCAL
//        cout<<"1"<<endl;
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int cas =1 ,_;
    cin>>_;
    while(_--)
    {
        run();
    }
    return 0;
}