hdu 1115 Lifting the Stone

　　题目链接：hdu 1115

　　计算几何求多边形的重心，弄清算法后就是裸题了，这儿有篇博客写得很不错的： 计算几何-多边形的重心

　　代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N = 1000006;

struct point {
    double x,y;
    point() {}
    point(double x, double y): x(x), y(y) {}
    void read()  {   scanf("%lf %lf",&x,&y);   }
    void readint() {
        int x,y;
        scanf("%d %d",&x,&y);
        this->x = x;
        this->y = y;
    }
    point operator - (const point &p2) const {
        return point(x - p2.x, y - p2.y);
    }
} p[N];

typedef point Vector;

double cross(Vector a, Vector b) {
    return a.x * b.y - a.y * b.x;
}

double Area(point a, point b, point c) {
    return cross(b - a, c - a) / 2;
}

inline double center3(double a, double b, double c) {
    return a + b + c;
}

int main() {
    int t,n;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i)
            p[i].read();

        double up = 0, down = 0;
        point ans;
        for(int i = 2; i <= n - 1; ++i) {
            up += center3(p[1].x, p[i].x, p[i + 1].x) * Area(p[1], p[i], p[i + 1]);
            down += Area(p[1], p[i], p[i + 1]);
        }
        ans.x = up / down / 3;

        up = 0;     down = 0;
        for(int i = 2; i <= n - 1; ++i) {
            up += center3(p[1].y, p[i].y, p[i + 1].y) * Area(p[1], p[i], p[i + 1]);
            down += Area(p[1], p[i], p[i + 1]);
        }
        ans.y = up / down / 3;
        printf("%.2lf %.2lf
", ans.x, ans.y);
    }
    return 0;
}

　　有个要注意的小细节，对每个小三角形求重心时对最后结果 / 3 即可，而不必在中间过程 / 3，应该是精度问题，所以除法应尽可能避免，因为这题的 n 有点大，所以不断 / 3 操作损失的精度是很大的。