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给个图，告诉R,P,Q三点的坐标，求出A,B,C三点的坐标

我的做法：

根据梅涅劳斯定理列出三个二元一次方程组，求出pb,qc,ra的长度，然后用点位移求出A,B,C三点的坐标即可

我的代码：

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
struct dot  
{  
    double x,y;  
    dot(){}  
    dot(double a,double b){x=a,y=b;}  
    friend dot operator -(dot a,dot b){return dot(a.x-b.x,a.y-b.y);}
    friend dot operator +(dot a,dot b){return dot(a.x+b.x,a.y+b.y);}
    friend dot operator *(dot a,double b){return dot(a.x*b,a.y*b);}
    friend double operator /(dot a,dot b){return a.x*b.x+a.y*b.y;}  
    friend double operator *(dot a,dot b){return a.x*b.y-a.y*b.x;}
};
struct fun  
{  
    double a,b,c;  
    fun(){}  
    fun(double x,double y,double z)  
    {  
        a=x;  
        b=y;  
        c=z;  
    }  
};  
dot sf(fun a,fun b)  
{  
    double c,d,e;  
    c=dot(a.a,b.a)*dot(a.b,b.b);  
    d=dot(a.c,b.c)*dot(a.b,b.b);  
    e=dot(a.a,b.a)*dot(a.c,b.c);  
    return dot(d/c,e/c);  
}
double dis(dot a,dot b){return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2));}
dot cd(dot a)
{
	double t=dis(a,dot(0,0));
	return dot(a.x/t,a.y/t);
}
int main()
{
	int N,i;
	double m[10],pr,pq,rq;
	dot a,b,c,d[10];
	cin>>N;
	while(N--)
	{
		for(i=0;i<3;i++)
			cin>>d[i].x>>d[i].y;
		for(i=1;i<7;i++)
			cin>>m[i];
		pr=dis(d[0],d[2]);
		pq=dis(d[0],d[1]);
		rq=dis(d[1],d[2]);
		a=sf(fun((m[1]+m[2])*m[4],-m[1]*m[3],m[1]*m[3]*pr),fun(m[5]*(m[1]+m[2]),-m[2]*m[6],-m[5]*(m[1]+m[2])*pr));
		a=cd(d[2]-d[0])*a.y+d[2];
		b=sf(fun(m[2]*m[4],-m[1]*(m[3]+m[4]),m[1]*(m[3]+m[4])*pq),fun(m[3]*m[5],-m[6]*(m[3]+m[4]),-m[3]*m[5]*pq));
		b=cd(d[0]-d[1])*b.x+d[0];
		c=sf(fun(m[3]*(m[5]+m[6]),-m[4]*m[6],-m[3]*(m[5]+m[6])*rq),fun(m[2]*(m[5]+m[6]),-m[1]*m[5],m[1]*m[5]*rq));
		c=cd(d[1]-d[2])*c.y+d[1];
		printf("%.8lf %.8lf %.8lf %.8lf %.8lf %.8lf\n",a.x,a.y,b.x,b.y,c.x,c.y);
	}
}

Time limit: 3.000 seconds

In the picture below you can see a triangle ABC.Point D, E and F divides the sides BC, CAand AB into m₁:m₂, m₃:m₄and m₅:m₆ ratios respectively. A, D; B,E and C, F are connected. AD and BE intersects at P,BE and CF intersects at Q and CF and ADintersects at R.

So now a new triangle PQR is formed. Given triangleABC it is very easy to find triangle PQR, but given triangle PQRit is not straight forward to find ABC. Your task is now to do that.

Input

First line of the input file contains an integer N (0< N < 25001) which denotes how many sets of inputs are there. Inputfor each set contains six floating-point number P_x, P_y,Q_x, Q_y, R_x, R_y. (0 ≤ P_x,P_y, Q_x, Q_y, R_x, R_y ≤10000) in one line and six positive integers m₁, m₂,m₃, m₄, m₅, m₆ (m₁<m₂,m₃<m₄ and m₅<m₆)in another line. These six numbers denote that the coordinate of points P, Qand R are (P_x, P_y), (Q_x, Q_y)and (R_x,R_y) respectively.P, Q and R will never be collinear and will be distinct and therewill always be a triangle ABC for the given input triangle PQR.Also note that P, Q and R will be given in counterclockwise order in the input.

Output

For each line of input produce one line of output. Thisline contains six floating-point numbers. These six integers denote the coordinatesof A, B and C. That is the first two integers denote thecoordinate of A, the third and fourth integers denote the coordinate of Band fifth and sixth integers denotes the coordinate of C. A, Band C will appear counter clockwise order. All the output numbers shouldhave eight digits after the decimal point.

Sample Input

Output for Sample Input

Problemsetter: Shahriar Manzoor, Special Thanks: Rujia Liu

UVA LIVE-4413 - Triangle Hazard

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原文地址：http://blog.csdn.net/stl112514/article/details/39315465