poj 1385 Lifting the Stone

题意：给出n个按构成简单多边形顺序的点,求出多边形的重心

简单多边形重心：sum（各个三角形重心 * 三角形有向面积权重）

各个三角形如图所示：
 

#include <iostream>#include <cstdio>#include <queue>#include <algorithm>#define MAX 1005#include <complex>#include <ctime>#include <cstdlib>#include <cstring>#include <string>#define INF 1e8#define MAX (int)1e6+ 5using namespace std;struct Point{    double x, y;    Point(){}    Point(const double &x, const double &y):x(x), y(y){}    Point operator + (const Point &b)const{        return  Point(x + b.x, y + b.y);    }    Point operator / (const double&b)const{        return  Point(x / b, y / b);    }    Point operator * (const double&b)const{        return  Point(x * b, y * b);    }    friend double dot(const Point &a, const Point &b){        return a.x * b.x + a.y * b.y;    }    friend double det(const Point &a, const Point &b){        return a.x * b.y - a.y * b.x;    }    void in(){        scanf("%lf %lf", &x, &y);    }    void out(){        printf("%.2f %.2f\n", x, y);    }};inline int next(const int &i, const int &n){ return (i + 1) % n;}//该简单多变性按构成多边形的顺序给出点。struct Polygon{    vector<Point>p;    Polygon(){}    Polygon(const vector<Point>pp):p(pp){}    //有向面积和,有可能为负,绝对值为多边形面积    double area(){        int n = p.size(), ni;        double ans  = 0;        for(int i = 0; i < n; i++ ){            ni  = next(i, n);            ans += det(p[i], p[ni]);    //逆时针        }        return ans / 2;    }    //取每个三角形的中心, 以有向面积作为权重, 他们的和就是重心    Point massCenter(){        Point center(0, 0);        int n = p.size(), ni;        for(int i = 0; i < n; i++){            ni = next(i, n);            center = center + (p[i] + p[ni]) * det(p[i], p[ni]);//逆时针        }        return center / area() / 6;    }};int T, n;int main(){    cin>>T;    double x, y;    while(T--){        cin>>n;        vector<Point>p;        for(int i = 0; i < n; i++){            scanf("%lf %lf", &x, &y);            p.push_back(Point(x, y));        }        Polygon(p).massCenter().out();    }    return 0;}