計算幾何總結

設平面上有N 個離散資料點( xi , yi ) ( i = 1, 2, ., n) , 其
多邊形重心G( . x1, . y1) 為:

　　成累加和:

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;struct Point{double x,y;};//叉積求面積double Area(const Point p0,const Point p1,const Point p2){    // 另外在求解的過程中，不需要考慮點的輸入順序是順時針還是逆時針，相除後就抵消了。     return (p0.x*p1.y+p1.x*p2.y+p2.x*p0.y-p1.x*p0.y-p2.x*p1.y-p0.x*p2.y)/2.0;}int main(){    int T,n;    scanf("%d",&T);    while(T--)    {        Point p0,p1,p2;        double area = 0,suma = 0,sumx = 0,sumy = 0;        scanf("%d",&n);        scanf("%lf%lf",&p0.x,&p0.y);        scanf("%lf%lf",&p1.x,&p1.y);        for(int i = 2;i < n;++i)        {            scanf("%lf%lf",&p2.x,&p2.y);            area = Area(p0,p1,p2);            suma += area;            sumx += (p0.x+p1.x+p2.x)*area;            sumy += (p0.y+p1.y+p2.y)*area;            p1 = p2;        }           printf("%.2lf %.2lf\n",sumx/suma/3,sumy/suma/3);    }    return 0;}