poj1266Cover an Arc（三角形外接圆）

链接

求出三角形的外接圆，通过圆心和半径可以知道这个圆的上下左右最远点，分别判断这个四个点跟弧的两端点A,B的关系，假如判断P点，弧内给出点为C，判断PC是否与AB相交即可判断出P是否在弧上。

精度问题 ceil-eps floor+eps

  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<stdlib.h>
  6 #include<vector>
  7 #include<cmath>
  8 #include<queue>
  9 #include<set>
 10 using namespace std;
 11 #define N 100000
 12 #define LL long long
 13 #define INF 0xfffffff
 14 const double eps = 1e-8;
 15 const double pi = acos(-1.0);
 16 const double inf = ~0u>>2;
 17 struct Point
 18 {
 19     double x,y;
 20     Point(double x=0,double y=0):x(x),y(y) {}
 21 }p[4];
 22 typedef Point pointt;
 23 pointt operator + (Point a,Point b)
 24 {
 25      return Point(a.x+b.x,a.y+b.y);
 26 }
 27 pointt operator - (Point a,Point b)
 28 {
 29         return Point(a.x-b.x,a.y-b.y);
 30 }
 31 int dcmp(double x)
 32 {
 33     if(fabs(x)<eps) return 0;
 34     else return x<0?-1:1;
 35 }
 36 struct Circle
 37 {
 38     Point center;
 39     double r;
 40 };
 41 double cross(Point a,Point b)
 42 {
 43     return a.x*b.y-a.y*b.x;
 44 }
 45 double mul(Point p0,Point p1,Point p2)
 46 {
 47     return cross(p1-p0,p2-p0);
 48 }
 49 double dis(Point a)
 50 {
 51     return a.x*a.x+a.y*a.y;
 52 }
 53 double area()
 54 {
 55     return fabs(cross(p[1]-p[3],p[2]-p[3]))/2;
 56 }
 57 bool seginter(pointt a1,pointt a2,pointt b1,pointt b2)
 58 {
 59      double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),
 60             c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);
 61      return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;
 62 }
 63 struct Circle Circumcircle()
 64 {
 65     Circle tmp;
 66     double a,b,c,c1,c2;
 67     double xa,ya,xb,yb,xc,yc;
 68     a = sqrt(dis(p[3]-p[1]));
 69     b = sqrt(dis(p[1]-p[2]));
 70     c = sqrt(dis(p[2]-p[3]));
 71     //根据s = a*b*c/R/4，求半径
 72     tmp.r = (a*b*c)/(area()*4.0);
 73     xa = p[3].x;
 74     ya = p[3].y;
 75     xb = p[1].x;
 76     yb = p[1].y;
 77     xc = p[2].x;
 78     yc = p[2].y;
 79     c1 = (dis(p[3])-dis(p[1]))/2;
 80     c2 = (dis(p[3])-dis(p[2]))/2;
 81     tmp.center.x = (c1*(ya-yc)-c2*(ya-yb))/((xa-xb)*(ya-yc)-(xa-xc)*(ya-yb));
 82     tmp.center.y = (c1*(xa-xc)-c2*(xa-xb))/((ya-yb)*(xa-xc)-(ya-yc)*(xa-xb));
 83     return tmp;
 84 }
 85 int main()
 86 {
 87     int i;
 88     double r;
 89     int w0,w1,h0,h1;
 90     for(i = 1; i <= 3 ; i++)
 91     scanf("%lf%lf",&p[i].x,&p[i].y);
 92     Circle cc = Circumcircle();
 93     r = cc.r;
 94
 95     Point q[5];
 96     for(i = 1 ;i < 5 ;i++)
 97     q[i] = cc.center;
 98     q[1].x-=r,q[2].x+=r,q[3].y-=r,q[4].y+=r;
 99
100     if(!seginter(q[1],p[3],p[1],p[2])) w0 = floor(q[1].x+eps);
101     else w0 = floor(min(p[1].x,p[2].x)+eps);
102
103     if(!seginter(q[2],p[3],p[1],p[2]))  w1 = ceil(q[2].x-eps);
104     else w1 = ceil(max(p[1].x,p[2].x)-eps);
105
106     if(!seginter(q[3],p[3],p[1],p[2])) h0 = floor(q[3].y+eps);
107     else h0 = floor(min(p[1].y,p[2].y)+eps);
108
109     if(!seginter(q[4],p[3],p[1],p[2])) h1 = ceil(q[4].y-eps);
110     else h1 = ceil(max(p[1].y,p[2].y)-eps);
111
112     printf("%d\n",(h1-h0)*(w1-w0));
113     return 0;
114 }

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