POJ

题目大意：给出一条圆弧上的两个端点A,B，和圆弧上两端点之间的一个点C，现在要用一块各个定点的坐标均为整数的矩形去覆盖这个圆弧，要求最小的矩形面积。

思路：叉积在本体发挥很强大的作用。首先求出三个点所在圆的圆心，也就是三角形的外心，然后判断着个圆上最上，最下，最左，最右四个点是否在该圆弧上，如果在，那么所求矩形的最上，最下，最左，最右边的坐标就是对应的点的坐标，否则，应该有圆弧两端点的坐标的相对大小来确定！

要判断一个点是否在圆弧上，主要用到叉积！把AB连结起来，设待检测的点式P，则如果是p在圆弧上，那么点p和点C在直线AB 的同一侧，否则在异侧，所以可由叉积判断！

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
  using namespace std;
  #define N 100000
  #define LL long long
  #define INF 0xfffffff
  const double eps = 1e-8;
  const double pi = acos(-1.0);
 const double inf = ~0u>>2;
  struct Point
 {
      double x,y;
      Point(double x=0,double y=0):x(x),y(y) {}
  }p[4];
  typedef Point pointt;
  pointt operator + (Point a,Point b)
  {
       return Point(a.x+b.x,a.y+b.y);
  }
  pointt operator - (Point a,Point b)
  {
         return Point(a.x-b.x,a.y-b.y);
  }
  int dcmp(double x)
  {
      if(fabs(x)<eps) return 0;
      else return x<0?-1:1;
  }
  struct Circle
  {
      Point center;
      double r;
  };
  double cross(Point a,Point b)
  {
      return a.x*b.y-a.y*b.x;
  }
  double mul(Point p0,Point p1,Point p2)
  {
      return cross(p1-p0,p2-p0);
  }
  double dis(Point a)
  {
      return a.x*a.x+a.y*a.y;
 }
  double area()
 {
      return fabs(cross(p[1]-p[3],p[2]-p[3]))/2;
  }
  bool seginter(pointt a1,pointt a2,pointt b1,pointt b2)
  {
       double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),
              c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);
       return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
  }
  struct Circle Circumcircle()
 {
      Circle tmp;
      double a,b,c,c1,c2;
     double xa,ya,xb,yb,xc,yc;
      a = sqrt(dis(p[3]-p[1]));
      b = sqrt(dis(p[1]-p[2]));
      c = sqrt(dis(p[2]-p[3]));
     tmp.r = (a*b*c)/(area()*4.0);
      xa = p[3].x;
      ya = p[3].y;
      xb = p[1].x;
      yb = p[1].y;
     xc = p[2].x;
      yc = p[2].y;
      c1 = (dis(p[3])-dis(p[1]))/2;
      c2 = (dis(p[3])-dis(p[2]))/2;
      tmp.center.x = (c1*(ya-yc)-c2*(ya-yb))/((xa-xb)*(ya-yc)-(xa-xc)*(ya-yb));
     tmp.center.y = (c1*(xa-xc)-c2*(xa-xb))/((ya-yb)*(xa-xc)-(ya-yc)*(xa-xb));
      return tmp;
  }
  int main()
  {
     int i;
      double r;
      int w0,w1,h0,h1;
      for(i = 1; i <= 3 ; i++)
      scanf("%lf%lf",&p[i].x,&p[i].y);
      Circle cc = Circumcircle();
      r = cc.r;

      Point q[5];
      for(i = 1 ;i < 5 ;i++)
     q[i] = cc.center;
      q[1].x-=r,q[2].x+=r,q[3].y-=r,q[4].y+=r;

    if(!seginter(q[1],p[3],p[1],p[2])) w0 = floor(q[1].x+eps);
     else w0 = floor(min(p[1].x,p[2].x)+eps);

     if(!seginter(q[2],p[3],p[1],p[2]))  w1 = ceil(q[2].x-eps);
    else w1 = ceil(max(p[1].x,p[2].x)-eps);

     if(!seginter(q[3],p[3],p[1],p[2])) h0 = floor(q[3].y+eps);
     else h0 = floor(min(p[1].y,p[2].y)+eps);

     if(!seginter(q[4],p[3],p[1],p[2])) h1 = ceil(q[4].y-eps);
     else h1 = ceil(max(p[1].y,p[2].y)-eps);

     printf("%d
",(h1-h0)*(w1-w0));
     return 0;
 }