求多边形的面积

http://acm.hdu.edu.cn/showproblem.php?pid=1115

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) 。该三角形的有向面积为:

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

　　△A1A2A3 边界构成逆时针回路时取+

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
struct point{
    double x,y;
};
double area(point p1,point p2,point p3)
{
    return ((p2.x-p1.x)*(p3.y-p1.y)-(p2.y-p1.y)*(p3.x-p1.x))/2;
}
int main()
{
    int n,t;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        point p1,p2,p3;
        double sum_x=0,sum_y=0,sum_area=0,s;
        scanf("%lf%lf",&p1.x,&p1.y);
        scanf("%lf%lf",&p2.x,&p2.y);
        for(int i=2;i<n;i++)
        {
            scanf("%lf%lf",&p3.x,&p3.y);
            s=area(p1,p2,p3);
            sum_area+=s;
            sum_x+=(p1.x+p2.x+p3.x)*s;
            sum_y+=(p1.y+p2.y+p3.y)*s;
            p2=p3;
        }
        printf("%.2lf %.2lf\n",sum_x/sum_area/3,sum_y/sum_area/3);
    }
    return 0;
}