UVA 12165 Triangle Hazard

Posted 2022-10-04 sahdsg

https://cn.vjudge.net/problem/UVA-12165

题目

 

给出D、E、F分BC，CA，AB的比$m_1:m_2$，$m_3:m_4$，$m_5:m_6$和PQR三点的坐标，求ABC三点的坐标

题解

利用梅涅劳斯定理，找出直线和三边的交点，然后每个边按顺序乘下去

可以写出三个方程

\\[\\fracARRP\\cdot\\boxed\\fracPQQB\\cdot\\fracBFFA=1\\]

\\[\\fracBPPQ\\cdot\\boxed\\fracQRRC\\cdot\\fracCDDB=1\\]

\\[\\fracCQQR\\cdot\\boxed\\fracRPPA\\cdot\\fracAEEC=1\\]

然后得到

\\[\\fracARRP\\cdot\\fracPQPQ+BP=\\fracm5m6=k_1\\]

\\[\\fracBPPQ\\cdot\\fracQRQR+RC=\\fracm1m2=k_2\\]

\\[\\fracCQQR\\cdot\\fracRPRP+RA=\\fracm3m4=k_3\\]

 

 最后解

\\[x_1=k_1(1+x_2)\\]

\\[x_2=k_2(1+x_3)\\]

\\[x_3=k_3(1+x_4)\\]

然后点加向量就可以得出三点坐标

AC代码

#include<cstdio>
#include<cctype>
#include<cmath>
#define REP(r,x,y) for(register int r=(x); r<(y);r++)
#ifdef sahdsg
#define DBG(...) printf(__VA_ARGS__)
#else
#define DBG(...) (void)0
#endif
using namespace std;

int _s; char _c;
template <class T>
inline void read(T&x) 
	x=0;
	do _c=getchar(); while(!isdigit(_c) && _c!=‘-‘);
	_s=1;
	if(_c==‘-‘) _s=-1, _c=getchar();
	while(isdigit(_c))  x=x*10+_c-‘0‘; _c=getchar(); x*=_s;

template<class T, class...A> inline void read(T &x, A&...a)read(x); read(a...);

#define D point
#define CD const D
struct point 
	double x,y;
	void read() scanf("%lf%lf",&x,&y);
	void prn() printf("%.8lf %.8lf",x,y);
	;
	D operator+(CD&l, CD&r) return (D)l.x+r.x,l.y+r.y;
	D operator-(CD&l, CD&r) return (D)l.x-r.x,l.y-r.y;
	D operator/(CD&l,double a) return (D)l.x/a,l.y/a;
	D operator*(CD&l,double a) return (D)l.x*a,l.y*a;
	D operator*(double a, CD &l) return (D)l.x*a,l.y*a;
	double cross(CD&l, CD&r) return l.x*r.y-l.y*r.x;
	D intersec(CD&a, D b, CD&c, D d) 
		b=b-a; d=d-c; D u=a-c;
		double t = cross(d, u) / cross(b,d);
		return a+b*t;
	
#undef CD
#undef D
point P,Q,R,A,B,C;
#define x1 nvdsaokvl
#define x2 nvkjdavnf
#define x3 vmasdvddz
int m1,m2,m3,m4,m5,m6;
double k1,k2,k3,x1,x2,x3;
int main() 
	int N; read(N);
	while(0<N--) 
		P.read(); Q.read(); R.read();
		read(m1,m2,m3,m4,m5,m6);
		k1=(double)m5/m6, k2=(double)m1/m2, k3=(double)m3/m4;
		double t=1-k1*k2*k3;
		x1=(k1+k1*k2*k3+k1*k2)/t;
		x2=(k2+k1*k2*k3+k2*k3)/t;
		x3=(k3+k1*k2*k3+k1*k3)/t;
		DBG("%lf %lf %lf\\n", x1,x2,x3);
		A=x1*(R-P)+R;
		B=x2*(P-Q)+P;
		C=x3*(Q-R)+Q;
		A.prn();putchar(‘ ‘);
		B.prn();putchar(‘ ‘);
		C.prn();putchar(‘\\n‘);
	
	return 0;