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：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<cmath> 5 using namespace std; 6 const int N = 1000006; 7 8 struct point { 9 double x,y; 10 point() {} 11 point(double x, double y): x(x), y(y) {} 12 void read() { scanf("%lf %lf",&x,&y); } 13 void readint() { 14 int x,y; 15 scanf("%d %d",&x,&y); 16 this->x = x; 17 this->y = y; 18 } 19 point operator - (const point &p2) const { 20 return point(x - p2.x, y - p2.y); 21 } 22 } p[N]; 23 24 typedef point Vector; 25 26 double cross(Vector a, Vector b) { 27 return a.x * b.y - a.y * b.x; 28 } 29 30 double Area(point a, point b, point c) { 31 return cross(b - a, c - a) / 2; 32 } 33 34 inline double center3(double a, double b, double c) { 35 return a + b + c; 36 } 37 38 int main() { 39 int t,n; 40 scanf("%d",&t); 41 while(t--) { 42 scanf("%d",&n); 43 for(int i = 1; i <= n; ++i) 44 p[i].read(); 45 46 double up = 0, down = 0; 47 point ans; 48 for(int i = 2; i <= n - 1; ++i) { 49 up += center3(p[1].x, p[i].x, p[i + 1].x) * Area(p[1], p[i], p[i + 1]); 50 down += Area(p[1], p[i], p[i + 1]); 51 } 52 ans.x = up / down / 3; 53 54 up = 0; down = 0; 55 for(int i = 2; i <= n - 1; ++i) { 56 up += center3(p[1].y, p[i].y, p[i + 1].y) * Area(p[1], p[i], p[i + 1]); 57 down += Area(p[1], p[i], p[i + 1]); 58 } 59 ans.y = up / down / 3; 60 printf("%.2lf %.2lf\n", ans.x, ans.y); 61 } 62 return 0; 63 }
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。
