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 }
poj1266Cover an Arc(三角形外接圆),布布扣,bubuko.com
原文:http://www.cnblogs.com/shangyu/p/3822486.html