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
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）,布布扣,bubuko.com
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）