首页 > 代码库 > 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）

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）

2024-07-02 21:43:32   227人阅读

 

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：
 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <cmath>
 4 using namespace std;
 5 #define eps 1e-10
 6 #define PI acos(-1)
 7 struct Point{
 8     double x,y;
 9     Point(double x=0,double y=0):x(x),y(y){}
10 };
11 typedef Point Vector ;
12 Vector operator + (Vector a,Vector b)
13 {
14     return Vector(a.x+b.x,a.y+b.y);
15 }
16 Vector operator - (Point a,Point b)
17 {
18     return Vector(a.x-b.x,a.y-b.y);
19 }
20 Vector operator * (Vector a,double b)
21 {
22     return Vector(a.x*b,a.y*b);
23 }
24 Vector operator / (Vector a,double b)
25 {
26     return Vector(a.x/b,a.y/b);
27 }
28 double Cross(Vector a,Vector b)
29 {
30     return a.x*b.y-b.x*a.y;
31 }
32 Vector Rotate(Vector A,double rad)
33 {
34     return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
35 }
36 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
37 {
38     Vector u = P-Q;
39     double t = Cross(w,u) / Cross(v,w);
40     return P+v*t;
41 }
42 int main()
43 {
44     int T;
45     scanf("%d",&T);
46     while(T--){
47         Point a,b;
48         scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
49         Point c = GetLineIntersection(a,Rotate(b-a,PI/3),b,Rotate(b-a,PI*2/3));
50         printf("(%.2lf,%.2lf)\n",c.x,c.y);
51     }
52     return 0;
53 }
54  
55 
56 
57 
58 /**************************************
59     Problem id    : SDUT OJ 2603 
60     User name    : Miracle 
61     Result        : Accepted 
62     Take Memory    : 512K 
63     Take Time    : 0MS 
64     Submit Time    : 2014-05-04 09:16:11  
65 **************************************/

 

Freecode : www.cnblogs.com/yym2013


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