POJ1266 Cover an Arc （计算几何模板）

Cover an Arc.

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 852 Accepted: 321

Description

A huge dancing-hall was constructed for the Ural State University's 80-th anniversary celebration. The size of the hall is 2000 * 2000 metres! The floor was made of square mirror plates with side equal to 1 metre. Then the walls were painted with an indelible paint. Unfortunately, in the end the painter flapped the brush and the beautiful mirror floor was stained with the paint. But not everything is lost yet! The stains can be covered with a carpet. 
Nobody knows why, but the paint on the floor formed an arc of a circle (a centre of the circle lies inside the hall). The dean of the Department of Mathematics and Mechanics measured the coordinates of the arc's ends and of some other point of the arc (he is sure that this information is quite enough for any student of the Ural State University). The dean wants to cover the arc with a rectangular carpet. The sides of a carpet must go along the sides of the mirror plates (so, the corners of the carpet must have integer coordinates). 
You should find the minimal square of such a carpet. 

Input

The input consists of six integers. At first the coordinates of the arc's ends are given. The co-ordinates of an inner point of the arc follow them. Absolute value of coordinates doesn't exceed 1000. The points don't belong the same straight line. The arc lies inside the square [-1000,1000] * [-1000,1000].

Output

You should write to the standard output the minimal square of the carpet covering this arc.

Sample Input

476 612487 615478 616

Sample Output

66

题意：

给三个点，求出能够覆盖由这三个点组成的圆弧的最小矩形面积。矩形必须符合题目要求，边为垂直或水平。

思路：

根据三个点可以组成一个三角形，那么就能算出这个三角形的外心。然后就能判断出是优弧还是劣弧，进行计算。

注意他的矩形要求是整数，所以要用ceil和floor进行取整。可能会有精度误差，可以考虑ceil时减去eps,floor时加上eps。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <string>#include <fstream>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;#define INF 0x3f3f3f3fconst int N=1005;const int mod=10000;const double eps=1e-8;struct Point{    double x,y;    Point(double x=0,double y=0):x(x),y(y){}};struct line{    Point a,b;    line(Point a=0,Point b=0):a(a),b(b){}};Point p[10005],ch[10005];line Line[5005];Point operator + (Point A ,Point B){    return Point(A.x+B.x,A.y+B.y);}Point operator - (Point A,Point B){    return Point(A.x-B.x,A.y-B.y);}Point operator * (Point A,double p){    return Point(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;    else if(x<0)return -1;    return 1;}bool operator ==(const Point &a,const Point &b){    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}double Dot(Point A,Point B){    return A.x*B.x+A.y*B.y;}double Cross(Point A,Point B){    return A.x*B.y-A.y*B.x;}double Length(Point A){    return sqrt(Dot(A,A));}double getdistance(Point P,line A){      //点到直线的距离    Point v1=A.b-A.a;    Point v2=P-A.a;    return fabs(Cross(v1,v2)/Length(v1));}double distances(Point p1,Point p2){  //两点距离    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}int  LeftofLine(Point P,line A){    //点在线的哪一侧，左侧为负    Point c1,c2;    c1=A.a;    c2=A.b;    double tmp=(c1.x-c2.x)/(c1.y-c2.y)*(P.y-c2.y)+c2.x;    if(tmp>P.x)return 1;         //left    else if(tmp<P.x)return 2;     //right    else return 4;}bool cmp(Point A ,Point B){    if(A.x==B.x)return A.y<B.y;    return A.x<B.x;}double Angle(Point A, Point B){        //两向量所成的夹角    return acos(Dot(A,B)/Length(A)/Length(B));}Point Rotate(Point A,double rad){     //向量逆时针旋转    return Point(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}Point GetLineIntersection(Point P,Point v,Point Q,Point w){    //两直线交点    Point u=P-Q;    double t=Cross(w,u)/Cross(v,w);    return P+v*t;}int ConvexHull(Point *p,int n,Point *ch){          //凸包    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-1])<=0) m--;        ch[m++]=p[i];    }    int k=m;    for(int i=n-1;i>=0;i--)    {        while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-1])<=0) m--;        ch[m++]=p[i];    }    if(n) m--;    return m;}bool SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2){  //线段规范相交    double c1=Cross(a2-a1,b1-a1);    double c2=Cross(a2-a1,b2-a1);    double c3=Cross(b2-b1,a1-b1);    double c4=Cross(b2-b1,a2-b1);    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;}Point intersection(line u,line v){     //两直线相交    Point ret=u.a;    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));    ret.x+=(u.b.x-u.a.x)*t;    ret.y+=(u.b.y-u.a.y)*t;    return ret;}Point circumcenter(Point a,Point b,Point c)  //三角形外心{    line u,v;    u.a.x=(a.x+b.x)/2;    u.a.y=(a.y+b.y)/2;    u.b.x=u.a.x-a.y+b.y;    u.b.y=u.a.y+a.x-b.x;    v.a.x=(a.x+c.x)/2;    v.a.y=(a.y+c.y)/2;    v.b.x=v.a.x-a.y+c.y;    v.b.y=v.a.y+a.x-c.x;    return intersection(u,v);}int main(){        int n;    int T,icase=0;    double ex1,ey1,ex2,ey2,xx,yy;    int S;    double dx,dy;    int mixi,miyi,maxi,mayi;    while(~scanf("%lf%lf%lf%lf%lf%lf",&ex1,&ey1,&ex2,&ey2,&xx,&yy)){        Point O,e1,e2,cc;        e1=Point(ex1,ey1);        e2=Point(ex2,ey2);        cc=Point(xx,yy);        O=circumcenter(e1,e2,cc);        double radius=distances(O,cc);        Point up,down,left,right;        up=Point(O.x,O.y+radius);        down=Point(O.x,O.y-radius);        left=Point(O.x-radius,O.y);        right=Point(O.x+radius,O.y);        //printf("%.0f %.0f %.0f %.0f\n",up.x,up.y,left.x,left.y);        int tt=0;        S=0;        if(SegmentProperIntersection(left,cc,e1,e2))mixi=floor(min(e1.x,e2.x));        else mixi=floor(left.x);                if(SegmentProperIntersection(up,cc,e1,e2))mayi=ceil(max(e1.y,e2.y));        else mayi=ceil(up.y);                if(SegmentProperIntersection(right,cc,e1,e2))maxi=ceil(max(e1.x,e2.x));        else maxi=ceil(right.x);                if(SegmentProperIntersection(down,cc,e1,e2))miyi=floor(min(e1.y,e2.y));        else miyi=floor(down.y);                printf("%d\n",(maxi-mixi)*(mayi-miyi));            }    }