POJ 3348 Cows (凸包面积)

来源：互联网发布：js实现分页思路编辑：程序博客网时间：2024/05/22 04:27

http://poj.org/problem?id=3348

水。

完整代码：

/*0ms,680KB*/#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const double eps = 1e-8;const int mx = 10005;struct point{double x, y;point(double x = 0.0, double y = 0.0): x(x), y(y) {}void read() const{scanf("%lf%lf", &x, &y);}bool operator == (const point& a) const{return fabs(x - a.x) < eps && fabs(y - a.y) < eps;}bool operator < (const point& a) const{return x < a.x || fabs(x - a.x) < eps && y < a.y;}};point a[mx], ans[mx];int n, len;///oa x obinline double cross_product(point& a, point& b){return a.x * b.y - a.y * b.x;}///ab x acinline double cross_product(point& a, point& b, point& c){return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);}///求凸包void convex_hull(){sort(a, a + n);//unique(a, a + n);len = 0;int i;for (i = 0; i < n; ++i){while (len >= 2 && cross_product(ans[len - 2], ans[len - 1], a[i]) < eps)--len;ans[len++] = a[i];}int tmp = len;for (i = n - 2; i >= 0; --i){while (len > tmp && cross_product(ans[len - 2], ans[len - 1], a[i]) < eps)--len;ans[len++] = a[i];}--len;}///求多边形面积 的两倍 。以原点为向量起点。注意ans[]要是顺时针顺序double area(){double sum = 0.0;ans[len] = ans[0];for (int i = 0; i < len; ++i) sum += cross_product(ans[i], ans[i + 1]);return sum;}int main(){scanf("%d", &n);for (int i = 0; i < n; ++i) a[i].read();if (n < 3) puts("0");else{convex_hull();printf("%d\n", (int)(area() / 100.0));}return 0;}

0 0