UVA 10652凸包+面的旋转

/*

UVA 10652计算几何
凸包+面的旋转
就凸包部分而言，是模板。
这道题有个小障碍，就是有确定的图形中心点和转动角度，确定转动后的顶点的坐标。
我的方法是，先转动中心点向顶点的向量，再中心加上这个向量。这应该是普适的。
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <set>
#include <algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-7
#define eps2 1e-3
#define zero(x) (((x)>0?(x):-(x))<eps)
using namespace std;


struct Point
{
    double x,y;
    Point() {}
    Point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
    bool operator<(const Point& p) const{
        if (x==p.x) return y<p.y;
        else return x<p.x;
    }
}P1[2505],P2[2505];

typedef Point Vector;

bool operator==(Point A,Point B)
{
    if ((fabs(A.x-B.x)<eps) && (fabs(A.y-B.y)<eps)) return true;
    else return false;
}
Vector operator-(Point A,Point B)//表示A指向B
{
    return Vector(A.x-B.x,A.y-B.y);
}
Vector operator*(Vector A,double k)
{
    return Vector(A.x*k,A.y*k);
}
Vector operator+(Point A,Point B)//表示A指向B
{
    return Vector(B.x+A.x,B.y+A.y);
}
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 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);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else if(x>0) return 1;
    else return -1;
}
double angle(Vector v)
{
    return atan2(v.y,v.x);
}
Vector Rotate(Vector A,double rad)//向量向逆时针旋转
{
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}

//p是原先点的数组，n是个数，ch是凸包的点集
//精度要求高是用dcmp比较
//返回凸包点的个数
int ConvexHull(Point *p, int n, Point* ch){        //求凸包
    sort(p, p + n);//先按照x，再按照y
    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;
}
double ConvexPolygonArea(Point *p, int n){//凸包面积
    double area = 0;
    for(int i = 1; i < n-1; i++) area += Area2(p[0], p[i], p[i+1]);
    return area / 2;
}

double x,y,w,h,arf;
int t,n,cnt1,cnt2;
int main()
{
    cin>>t;
    while(t--)
    {
        double area1=0,area2=0;
        cnt1=0;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>x>>y>>w>>h>>arf;
            arf=arf/180*M_PI;
            area1+=w*h;
            Vector v[4];//四个方向上的向量
            v[0].x=w/2;v[0].y=h/2;
            v[1].x=-w/2;v[1].y=h/2;
            v[2].x=-w/2;v[2].y=-h/2;
            v[3].x=w/2;v[3].y=-h/2;
            for(int j=0;j<4;j++)//构造4个顶点
            {
                Vector nv=Rotate(v[j],2*M_PI-arf);
                P1[cnt1].x=nv.x+x;
                P1[cnt1++].y=nv.y+y;
            }
        }
        sort(P1,P1+cnt1);
        int cnt2=ConvexHull(P1,cnt1,P2);
        area2=ConvexPolygonArea(P2,cnt2);
        double ans=area1/area2*100;
        printf("%.1lf %%
",ans);
    }
    return 0;
}