hdu - 4709 - Herding
题意:给出N个点的坐标,从中取些点来组成一个多边形,求这个多边形的最小面积,组不成多边形的输出"Impossible"(测试组数 T <= 25, 1 <= N <= 100, -1000 <= 坐标Xi, Yi <= 1000)。——>>面积最小,若有的话,一定是三角形。判断3点是否能组成一个三角形,若用斜率来做,麻烦且可能会有精度误差,用叉积来判断甚好(只需判断两向量的叉积是否为0)。
注意:N可为1、2,这时不能判断三角形。
#include <cstdio> #include <cmath> #include <algorithm> using namespace std; const int maxn = 100 + 10; const double eps = 1e-10; const double INF = 1 << 30; int N; struct Point{ double x; double y; Point(double x = 0, double y = 0):x(x), y(y){} }p[maxn]; typedef Point Vector; Vector operator + (Point A, Point B){ return Vector(A.x + B.x, A.y + B.y); } Vector operator - (Point A, Point B){ return Vector(A.x - B.x, A.y - B.y); } Vector operator * (Point A, double p){ return Vector(A.x * p, A.y * p); } Vector operator / (Point A, double p){ return Vector(A.x / p, A.y / p); } double Cross(Vector A, Vector B){ return A.x * B.y - B.x * A.y; } double Area2(Point A, Point B, Point C){ return Cross(B-A, C-A); } int dcmp(double x){ if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } void read(){ scanf("%d", &N); for(int i = 0; i < N; i++) scanf("%lf%lf", &p[i].x, &p[i].y); } void solve(){ double Min = INF; if(N >= 3){ for(int i = 0; i < N; i++) for(int j = i+1; j < N; j++) for(int k = j+1; k < N; k++) if(dcmp(Cross(p[j] - p[i], p[k] - p[i]))){ double temp = fabs(Area2(p[i], p[j], p[k])); Min = min(Min, temp); } } if(dcmp(Min - INF) == 0) puts("Impossible"); else printf("%.2f\n", Min/2); } int main() { int T; scanf("%d", &T); while(T--){ read(); solve(); } return 0; }
补充:软件开发 , C++ ,