sdut 2603:Rescue The Princess（第四届山东省省赛原题，计算几何，向量旋转 + 向量交点）

Rescue The Princess

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue the princess, a prince who wanted to marry the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.

    Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x₁,y₁) and B(x₂,y₂). The third coordinate C(x₃,y₃) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x₁,y₁) and B(x₂,y₂), but he doesn’t know where the C(x₃,y₃) is. The prince need your help. Can you calculate the C(x₃,y₃) and tell him?

输入

    The first line is an integer T(1 <= T <= 100) which is the number of test cases. T test cases follow. Each test case contains two coordinates A(x₁,y₁) and B(x₂,y₂), described by four floating-point numbers x₁, y₁, x₂, y₂ ( |x₁|, |y₁|, |x₂|, |y₂| <= 1000.0).
    Please notice that A(x₁,y₁) and B(x₂,y₂) and C(x₃,y₃) are in an anticlockwise direction from the equilateral triangle. And coordinates A(x₁,y₁) and B(x₂,y₂) are given by anticlockwise.

输出

    For each test case, you should output the coordinate of C(x₃,y₃), the result should be rounded to 2 decimal places in a line.

示例输入

4
-100.00 0.00 0.00 0.00
0.00 0.00 0.00 100.00
0.00 0.00 100.00 100.00
1.00 0.00 1.866 0.50

示例输出

(-50.00,86.60)
(-86.60,50.00)
(-36.60,136.60)
(1.00,1.00)

提示

 

来源

2013年山东省第四届ACM大学生程序设计竞赛

---

 

　　计算几何，向量旋转 + 向量交点。

　　这是一道水题，计算几何和解析几何的方法都可以做出来。

　　题意：abc是一个等边三角形，已知a、b两点坐标，且为逆时针方向，让你求点c的坐标。

　　思路：求出向量ab，然后分别逆时针旋转60°和120°求出ab为起点的两边的向量，最后求出这两条向量的交点。

　　代码：

#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;
#define eps 1e-10
#define PI acos(-1)
struct Point{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){}
};
typedef Point Vector ;
Vector operator + (Vector a,Vector b)
{
    return Vector(a.x+b.x,a.y+b.y);
}
Vector operator - (Point a,Point b)
{
    return Vector(a.x-b.x,a.y-b.y);
}
Vector operator * (Vector a,double b)
{
    return Vector(a.x*b,a.y*b);
}
Vector operator / (Vector a,double b)
{
    return Vector(a.x/b,a.y/b);
}
double Cross(Vector a,Vector b)
{
    return a.x*b.y-b.x*a.y;
}
Vector Rotate(Vector A,double rad)
{
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
{
    Vector u = P-Q;
    double t = Cross(w,u) / Cross(v,w);
    return P+v*t;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        Point a,b;
        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
        Point c = GetLineIntersection(a,Rotate(b-a,PI/3),b,Rotate(b-a,PI*2/3));
        printf("(%.2lf,%.2lf)
",c.x,c.y);
    }
    return 0;
}
 



/**************************************
    Problem id    : SDUT OJ 2603 
    User name    : Miracle 
    Result        : Accepted 
    Take Memory    : 512K 
    Take Time    : 0MS 
    Submit Time    : 2014-05-04 09:16:11  
**************************************/

Freecode : www.cnblogs.com/yym2013