uva 10652 Board Wrapping 凸包

题意:

给出一些矩形，(长、宽、中心位置、顺时针旋转的角度)，用一个最小的凸多边形把它围起来，求矩形面积和/凸多边形面积。

解法：求凸包

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])#define mem(a,x)  memset(a,x,sizeof a)typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;const double PI=cos(-1.0);struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y) {};};typedef Point Vector;Vector operator +(Vector A,Vector B) {return Vector(A.x+B.x,A.y+B.y); }Vector operator -(Vector A,Vector 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); }Vector operator -(Vector A)  {return  Vector(-A.x,-A.y);}double torad(double deg){    return deg/180*acos(-1);}const double eps=1e-10;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.x-b.x)==0 &&dcmp(a.y-b.y)<0;}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)//点乘{    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)//叉乘{    return A.x*B.y-A.y*B.x;}double Area2(Point A,Point B,Point C)//平行四边形面积{    return Cross(B-A,C-A);}Vector Rotate(Vector A,double rad)//向量A 逆时针旋转rad度{   return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad)     );}Vector Normal(Vector A)//计算A的单位向量{    double L=Length(A);    return Vector( -A.y/L,A.x/L );}int np;Point p[2500],ch[2500];int ConvexHell(Point *p,int n,Point *ch){    sort(p,p+n);    int m=0;    for(int i=0;i<n;i++)    {        while(m>1&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]   )) <=0   )  m--;//叉乘<=0，逆时针，上凸        ch[m++]=p[i];     }    int k=m;    for(int i=n-2;i>=0;i--)    {        while(m>k &&dcmp( 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(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;}int main(){   std::ios::sync_with_stdio(false);   int n,T;cin>>T;   while(T--)   {      cin>>n;      double x,y,w,h,ang,area1=0;      np=0;      for0(i,n)      {          cin>>x>>y>>w>>h>>ang;          Point o(x,y);          ang=-torad(ang);         p[np++]=o+Rotate(Vector(-w/2,-h/2),ang);         p[np++]=o+Rotate(Vector(w/2,-h/2),ang);         p[np++]=o+Rotate(Vector(-w/2,h/2),ang);         p[np++]=o+Rotate(Vector(w/2,h/2),ang);         area1+=w*h;      }      int m=ConvexHell(p,np,ch);      double area2=PolygonArea(ch, m);//      cout<<area1<<" "<<area2<<endl;      printf("%.1f %%\n",area1*100/area2);   }   return 0;}