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】：

 #include <iostream>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include<stdio.h>

 using namespace std;
 //定义点
 struct point
 {
     double x,y;
 };

 typedef struct point point;

 double fabs(double x)
 {
     return x<?-x:x;
 }
 //定义直线
 struct line
 {
     point a,b;
 };
 typedef struct line line;

 //两点距离
 double distance(point p1,point p2)
 {
     return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
 }
 //两直线求交点
 point intersection(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 incenter(point a,point b,point c)
 {
     line u,v;
     double m,n;
     u.a=a;
     m=atan2(b.y-a.y,b.x-a.x);
     n=atan2(c.y-a.y,c.x-a.x);
     u.b.x=u.a.x+cos((m+n)/);
     u.b.y=u.a.y+sin((m+n)/);
     v.a=b;
     m=atan2(a.y-b.y,a.x-b.x);
     n=atan2(c.y-b.y,c.x-b.x);
     v.b.x=v.a.x+cos((m+n)/);
     v.b.y=v.a.y+sin((m+n)/);
     return intersection(u,v);
 }

 int main()
 {
     point a,b,c;
     scanf("%lf%lf",&a.x,&a.y);
     scanf("%lf%lf",&b.x,&b.y);
     scanf("%lf%lf",&c.x,&c.y);
     point d;
     d = incenter(a,b,c);
     if(fabs(d.x)<1e-) d.x=;
     if(fabs(d.y)<1e-) d.y=;
     printf("%.2lf %.2lf\n",d.x,d.y);
     return ;
 }

【code2】：

 #include <iostream>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include<stdio.h>

 //using namespace std;
 //定义点
 struct point
 {
     double x,y;
 };

 double distance(point p1,point p2)
 {
     return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
 }

 int main()
 {
     point a,b,c;
     scanf("%lf%lf",&a.x,&a.y);
     scanf("%lf%lf",&b.x,&b.y);
     scanf("%lf%lf",&c.x,&c.y);
     point d;
     double ab = distance(a,b);
     double bc = distance(b,c);
     double ac = distance(a,c);
     d.x = (a.x*bc+b.x*ac+c.x*ab)/(ab+bc+ac);
     d.y = (a.y*bc+b.y*ac+c.y*ab)/(ab+bc+ac);
     printf("%.2lf %.2lf\n",d.x,d.y);
     return ;
 }