题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=32286
【思路】
凸包
根据角度与中心点求出长方形所有点来,然后就可以应用凸包算法了。
【代码】
#include<cmath> #include<cstdio> #include<algorithm> using namespace std; const double PI = acos(-1.0); double torad(double deg) { return deg/180 * PI; } //角度化弧度 struct Pt { double x,y; Pt(double x=0,double y=0):x(x),y(y) {}; }; typedef Pt vec; vec operator - (Pt A,Pt B) { return vec(A.x-B.x,A.y-B.y); } vec operator + (vec A,vec B) { return vec(A.x+B.x,A.y+B.y); } bool operator < (const Pt& a,const Pt& b) { return a.x<b.x || (a.x==b.x && a.y<b.y); } double cross(Pt A,Pt B) { return A.x*B.y-A.y*B.x; } vec rotate(vec A,double rad) { return vec(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)); } int ConvexHull(Pt* p,int n,Pt* ch) { sort(p,p+n); 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--; //维护凸包 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; } double PolygonArea(Pt* p,int n) { //多边形面积 double S=0; for(int i=1;i<n-1;i++) S += cross(p[i]-p[0],p[i+1]-p[0]); return S/2; } const int N = 2500+10; Pt P[N],ch[N]; int n; int main() { int T; scanf("%d",&T); while(T--) { scanf("%d",&n); int pc=0; double S1=0; double x,y,w,h,j; for(int i=0;i<n;i++) { scanf("%lf%lf%lf%lf%lf",&x,&y,&w,&h,&j); double ang=-torad(j); Pt o(x,y); P[pc++]= o + rotate(vec(-w/2,-h/2),ang); P[pc++]= o + rotate(vec(w/2,-h/2),ang); P[pc++]= o + rotate(vec(-w/2,h/2),ang); P[pc++]= o + rotate(vec(w/2,h/2),ang); S1 += w*h; } int m=ConvexHull(P,pc,ch); double S2=PolygonArea(ch,m); printf("%.1lf %%\n",S1*100/S2); } return 0; }
时间: 2024-10-04 07:33:14