大意:有n块矩形木板,你的任务是用一个面积尽量小的凸多边形把它们包起来,并计算出木板站整个包装面积的百分比。
思路:按照题意将所有矩形顶点坐标存起来,旋转时先旋转从中心出发的向量,求得各个坐标之后,求凸包即可。
#include <iostream> #include <cstdlib> #include <cstdio> #include <string> #include <cstring> #include <cmath> #include <vector> #include <queue> #include <stack> #include <algorithm> using namespace std; const double eps = 1e-10; const double PI = acos(-1.0); struct Point { double x, y; Point(double x = 0, double y = 0) : x(x), y(y) { } bool operator < (const Point& a) const { if(a.x != x) return x < a.x; return y < a.y; } }; typedef Point Vector; Vector operator + (Vector A, Vector 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 * (Vector A, double p) { return Vector(A.x*p, A.y*p); } Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; } bool operator == (const Point& a, const Point &b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); } double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); } Vector Rotate(Vector A, double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad)); } Point GetIntersection(Point P, Vector v, Point Q, Vector w) { Vector u = P-Q; double t = Cross(w, u) / Cross(v, w); return P+v*t; } bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2) { double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1); double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1); return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; } double PolygonArea(Point* p, int n) { double area = 0; for(int i = 1; i < n-1; i++) area += Cross(p[i]-p[0], p[i+1]-p[0]); return area/2; } bool OnSegment(Point p, Point a1, Point a2) { return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0; } double DistanceToLine(Point P, Point A, Point B) { Vector v1 = B-A, v2 = P-A; return fabs(Cross(v1, v2)) / Length(v1); } double DistanceToSegment(Point P, Point A, Point B) { if(A == B) return Length(P-A); Vector v1 = B-A, v2 = P-A, v3 = P-B; if(dcmp(Dot(v1, v2) < 0)) return Length(v2); else if(dcmp(Dot(v1, v3) > 0)) return Length(v3); else return fabs(Cross(v1, v2)) / Length(v1); } int ConvexHull(Point *p, int n, Point *ch) //凸包 { sort(p, p+n); n = unique(p, p+n) - p; //去重 int m = 0; for(int i = 0; i < n; i++) { while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; ch[m++] = p[i]; } int k = m; for(int i = n-2; i >= 0; i--) { while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--; ch[m++] = p[i]; } if(n > 1) m--; return m; } Point read_point() { Point A; scanf("%lf%lf", &A.x, &A.y); return A; } int n; int pc; double x, y, w, h, j, rad; double area1, area2; const int maxn = 2410; void init() { pc = 0; area1 = area2 = 0; } double torad(double n) //返回弧度 { return n / 180 * PI; } Point P[maxn], Q[maxn]; void read_case() { init(); scanf("%d", &n); for(int i = 0; i < n; i++) { scanf("%lf%lf%lf%lf%lf", &x, &y, &w, &h, &j); Point o(x, y); rad = -torad(j); P[pc++] = o + Rotate(Vector(-w/2, -h/2), rad); /*按逆时针方向把矩形的四个点分别顺时针旋转*/ P[pc++] = o + Rotate(Vector(w/2, -h/2), rad); P[pc++] = o + Rotate(Vector(w/2, h/2), rad); P[pc++] = o + Rotate(Vector(-w/2, h/2), rad); area1 += w*h; } } void solve() { read_case(); int m = ConvexHull(P, pc, Q); area2 = PolygonArea(Q, m); printf("%.1lf %%\n", area1*100/area2); } int main() { int T; scanf("%d", &T); while(T--) { solve(); } return 0; }