bnuoj 4209 Triangle（计算几何）
http://www.bnuoj.com/bnuoj/problem_show.php?pid=4209
题意：如题
题解：公式直接计算，或者角平分线求交点
【code1】：
1 #include <iostream> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include<stdio.h> 6 7 using namespace std; 8 //定义点 9 struct point 10 { 11 double x,y; 12 }; 13 14 typedef struct point point; 15 16 17 double fabs(double x) 18 { 19 return x<0?-x:x; 20 } 21 //定义直线 22 struct line 23 { 24 point a,b; 25 }; 26 typedef struct line line; 27 28 //两点距离 29 double distance(point p1,point p2) 30 { 31 return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); 32 } 33 //两直线求交点 34 point intersection(line u,line v) 35 { 36 point ret=u.a; 37 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)) 38 /((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x)); 39 ret.x+=(u.b.x-u.a.x)*t; 40 ret.y+=(u.b.y-u.a.y)*t; 41 return ret; 42 } 43 44 point incenter(point a,point b,point c) 45 { 46 line u,v; 47 double m,n; 48 u.a=a; 49 m=atan2(b.y-a.y,b.x-a.x); 50 n=atan2(c.y-a.y,c.x-a.x); 51 u.b.x=u.a.x+cos((m+n)/2); 52 u.b.y=u.a.y+sin((m+n)/2); 53 v.a=b; 54 m=atan2(a.y-b.y,a.x-b.x); 55 n=atan2(c.y-b.y,c.x-b.x); 56 v.b.x=v.a.x+cos((m+n)/2); 57 v.b.y=v.a.y+sin((m+n)/2); 58 return intersection(u,v); 59 } 60 61 int main() 62 { 63 point a,b,c; 64 scanf("%lf%lf",&a.x,&a.y); 65 scanf("%lf%lf",&b.x,&b.y); 66 scanf("%lf%lf",&c.x,&c.y); 67 point d; 68 d = incenter(a,b,c); 69 if(fabs(d.x)<1e-6) d.x=0; 70 if(fabs(d.y)<1e-6) d.y=0; 71 printf("%.2lf %.2lf\n",d.x,d.y); 72 return 0; 73 }
【code2】：
1 #include <iostream> 2 #include <math.h> 3 #include <string.h> 4 #include <stdlib.h> 5 #include<stdio.h> 6 7 //using namespace std; 8 //定义点 9 struct point 10 { 11 double x,y; 12 }; 13 14 double distance(point p1,point p2) 15 { 16 return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)); 17 } 18 19 int main() 20 { 21 point a,b,c; 22 scanf("%lf%lf",&a.x,&a.y); 23 scanf("%lf%lf",&b.x,&b.y); 24 scanf("%lf%lf",&c.x,&c.y); 25 point d; 26 double ab = distance(a,b); 27 double bc = distance(b,c); 28 double ac = distance(a,c); 29 d.x = (a.x*bc+b.x*ac+c.x*ab)/(ab+bc+ac); 30 d.y = (a.y*bc+b.y*ac+c.y*ab)/(ab+bc+ac); 31 printf("%.2lf %.2lf\n",d.x,d.y); 32 return 0; 33 }
