poj 1266 Cover an Arc.

http://poj.org/problem?id=1266

Cover an Arc.

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 823		Accepted: 308

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

476 612
487 615
478 616

Sample Output

66

Source

Ural State University Internal Contest October'2000 Students Session (URAL 1043)

 

 

 

分析：

几何题， 求正方形覆盖圆弧的面积。

 

 

 

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a
#include<math.h>
using namespace std;
#define eps 1e-8
struct point{double x,y;};
struct line {point a,b;};
point a,b,c;
double xmult(point p1,point p2,point p0){
 return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool pp(point p)
{
    double t1,t2;
    t1=(xmult(a,c,b));
    t2=(xmult(a,p,b));
    if ((t1<0&&t2<0)||(t1>0&&t2>0)) return true;
    return false;
}
double distan (point p1,point p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
point inter(line u,line v)
{
    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 circle(point a,point b,point c )
{
    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 inter(u,v);
}
int main()
{
    point d,e,p;
    int cas =1;
    while(~scanf("%lf %lf %lf %lf %lf %lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y))
    {
        d = circle(a,b,c);
        double bj = distan(d,a);
        double maxx,maxy,minx,miny;
        double dd=d.x,yy=d.y;
        int ax,bx,cx,ay,by,cy;
        maxx=max(a.x,b.x);
        maxx=max(maxx,c.x);
        minx=min(a.x,b.x);
        minx=min(minx,c.x);
        maxy=max(a.y,b.y);
        maxy=max(maxy,c.y);
        miny=min(a.y,b.y);
        miny=min(miny,c.y);
        p.x=d.x-bj;
        p.y=d.y;
        if(pp(p))
            minx=p.x;
        p.x=d.x+bj;
        if(pp(p))
            maxx=p.x;
        p.x=d.x;
        p.y=d.y-bj;
        if(pp(p))
            miny=p.y;
        p.y=d.y+bj;
        if(pp(p))
            maxy=p.y;
        cx=(long)ceil(maxx-eps)-(long)floor(minx+eps);
        cy=(long)ceil(maxy-eps)-(long)floor(miny+eps);
        printf("%d\n",cx*cy);
    }
    return 0;
}