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<0?-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)/2);
    u.b.y=u.a.y+sin((m+n)/2);
    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)/2);
    v.b.y=v.a.y+sin((m+n)/2);
    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-6) d.x=0;
    if(fabs(d.y)<1e-6) d.y=0;
    printf("%.2lf %.2lf
",d.x,d.y);
    return 0;
}

【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
",d.x,d.y);
    return 0;
}