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 ;

 }