uvalive 7366(二分 + 两向量的交点 + 向量的旋转)

题意：给出一个三角形的三个点，然后求使得图中的三个角相等的三角形内的点的坐标。

思路：因为有单调性，所以对角进行二分，然后用两个角求出第三个角，看是否符合。



#include<bits/stdc++.h>
using namespace std;
#define PI acos(-1.0)

typedef struct Node
{
double x,y;
Node operator - (Node t)
{
return (Node){x - t.x,y - t.y};
}
Node operator + (Node t)
{
return (Node){x + t.x,y + t.y};
}
//点积
double operator * (Node t)
{
return x * t.x + y * t.y;
}
//叉积
double operator ^(Node t)
{
return x * t.y - y*t.x;
}
Node operator * (double t)
{
return (Node){x * t,y * t};
}
}Vector;

double Angle(Vector p1,Vector p2)
{
return acos( (p1 * p2) / sqrt(p1 *p1) / sqrt(p2 * p2));
}
Vector Lrotate(Vector p,double angle)//逆时针旋转
{
return (Vector){p.x * cos(angle) - p.y * sin(angle) ,p.x * sin(angle) + p.y * cos(angle)};
}
Vector Rrotate(Vector p,double angle)//顺时针旋转
{
return (Vector){p.y * sin(angle) - p.x * cos(angle) ,p.x * sin(angle) + p.y * cos(angle)};
}

Node GetIn(Node P,Vector v,Node Q,Vector w) //两直线的交点
{
Vector u = P - Q;
double t = (w^u)/(v^w);
return P+v*t;
}
Node a[3];
Node s;

double get(double angle)
{
Vector p1 = Lrotate(a[2] - a[1],angle),p2 = Lrotate(a[1] - a[0],angle);
s = GetIn(a[0],p2,a[1],p1);
return Angle(s - a[2],a[0] - a[2]);
}

int main()
{
// freopen("in.txt","r",stdin);
int Tcase;scanf("%d",&Tcase);
while(Tcase --)
{
int k;scanf("%d",&k);
for(int i = 0; i < 3;i ++)scanf("%lf%lf",&a[i].x,&a[i].y);
double l = 0,r = PI/2;
r = min(r,Angle(a[1] - a[0],a[2] - a[0]));
r = min(r,Angle(a[2] - a[1],a[0] - a[1]));
r = min(r,Angle(a[0] - a[2],a[1] - a[2]));
while(r - l > 1e-10)
{
double mid = (l + r) / 2;
if(get(mid) > mid) l = mid;
else r = mid;
}
printf("%d %.5f %.5f\n",k,s.x,s.y);
}
return 0;
}