山東省第四屆ACM大學生程式設計競賽-Rescue The Princess（計算幾何）

kewlgrl發表於2016-04-09
ACM程式設計

Rescue The Princess

Time Limit: 1000ms   Memory limit: 65536K  有疑問？點這裡^_^

題目描述

    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 equilateraltriangle 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 youcalculate the C(x3,y3) and tell him?

輸入

    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 equilateraltriangle. And coordinates A(x1,y1) and B(x2,y2) are given by anticlockwise.

輸出

    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.

示例輸入

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

示例輸出

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

提示

 

來源

2013年山東省第四屆ACM大學生程式設計競賽

題目意思：
給定A、B兩個點的座標，求另外一個點C，使得ABC構成一個等邊三角形，且ABC三點是逆時針順序。

解題思路如圖（From 隊友大神CXF）：



注意到數學反三角函式double atan2 (double, double); 反正切(整圓值), 結果介於[-PI/2, PI/2]
區別於double atan (double); 反正切(主值), 結果介於[-PI/2, PI/2]，因為斜率為零的情況atan2不返回0而不是錯誤值。

/* 
* Copyright (c) 2016, 煙臺大學計算機與控制工程學院 
* All rights reserved. 
* 檔名稱：cal.cpp 
* 作    者：單昕昕 
* 完成日期：2016年4月9日 
* 版 本 號：v1.0 
*/  
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
const double PI=acos(-1.0);

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double x1,y1,x2,y2;
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        double l=sqrt((y1-y2)*(y1-y2)+(x1-x2)*(x1-x2));
        double s=atan2(y2-y1,x2-x1);
        double l1=l*sin(s+(PI/3.0));
        double l2=l*cos(s+(PI/3.0));
        double x=x1+l2;
        double y=y1+l1;
        printf("(%.2lf,%.2lf)\n",x,y);
    }
    return 0;
}



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