Ural 1043 Cover the Arc

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1043

题目大意：一个2000*2000方格坐标，x,y范围都是【-1000,1000】。现在给你一个圆弧，告诉你圆弧的两个端点和任意一个中间点。现在要你算出最小的矩形（长和宽都要为整数，即四个顶点在方格顶点上）来完全覆盖这个圆弧。

 算法思路：很明显要算出圆心，这个可以有线段中垂线交求，也可以由方程，只是很麻烦。然后以圆心找出这个圆的左右上下四个极点，判断是否在圆弧上（用叉积即可），与给出的三个点一起维护这段圆弧的四个方向的极大点。然后向上向下取整即可。

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;

const double eps = 1e-3;
const double INF = 1000000000.00;

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); }

bool operator < (const Point& A, const Point& B){
    return A.x < B.x || (A.x == B.x && A.y < B.y);
}
int dcmp(double x){
    if(fabs(x) < eps) return 0;
    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 Cross(Vector A,Vector B) { return A.x*B.y-A.y*B.x; }


Point read_point()
{
    Point P;
    scanf("%lf %lf",&P.x,&P.y);
    return P;
}
Point normal(Vector A) { return Vector(-A.y,A.x); };

Point GetLineIntersecion(Point P, Vector v,Point Q,Vector w){
    Vector u = P - Q;
    double t = Cross(w,u)/Cross(v,w);
    return P + v*t;
}

void Judge(Point& Pl,Point& Pr,Point& Pu,Point& Pd,Point P){
    if(Pl.x > P.x) Pl = P;
    if(Pr.x < P.x) Pr = P;
    if(Pu.y < P.y) Pu = P;
    if(Pd.y > P.y) Pd = P;
}

int main()
{
    //freopen("E:\acm\input.txt","r",stdin);
    Point Ps,Pt,Pi;
    Ps = read_point();
    Pt = read_point();
    Pi = read_point();
    Point mid1,mid2,O;

    mid1 = (Ps+Pi)/2;
    mid2 = (Pi+Pt)/2;
    O = GetLineIntersecion(mid1,normal(Ps-Pi),mid2,normal(Pi-Pt));

    double r = Length(O-Ps); 
    Point Pu,Pl,Pr,Pd;
    Pu = Pl = Pr = Pd = Ps;

    Judge(Pl,Pr,Pu,Pd,Pt);
    Judge(Pl,Pr,Pu,Pd,Pi);

    Point P = Point(O.x+r,O.y);
    if(dcmp(Cross(Pt-Ps,Pi-Ps)*Cross(Pt-Ps,P-Ps)) > 0){
        Judge(Pl,Pr,Pu,Pd,P);
    }
    P = Point(O.x-r,O.y);
    if(dcmp(Cross(Pt-Ps,Pi-Ps)*Cross(Pt-Ps,P-Ps)) > 0){
        Judge(Pl,Pr,Pu,Pd,P);
    }
    P = Point(O.x,O.y+r);
    if(dcmp(Cross(Pt-Ps,Pi-Ps)*Cross(Pt-Ps,P-Ps)) > 0){
        Judge(Pl,Pr,Pu,Pd,P);
    }
    P = Point(O.x,O.y-r);
    if(dcmp(Cross(Pt-Ps,Pi-Ps)*Cross(Pt-Ps,P-Ps)) > 0){
        Judge(Pl,Pr,Pu,Pd,P);
    }
    int width = ceil(Pu.y-eps) - floor(Pd.y+eps); 
    int length = ceil(Pr.x-eps) - floor(Pl.x+eps);
    printf("%d
",width*length);
}