[Training] UVA 1595 Symmetry

本題為 Brute Force 可以迅速解題。

題目給予一些二維座標，求是否所有點左右對稱。

看似簡單，但需要注意以下幾點，不然很容易拖掉 Debug 時間。

1.   scanf 讀取輸入的時候，遇到自定義的 struct，不必遲疑，直接把 “&” 寫在前面：

         

    scanf("%d %d", &a[i].x, &a[i].y);

2.   取 x 最大值和最小值，相加當作中點，不需要除以 2，避免出現浮點數問題。

3.   x_min 和 x_max 取完，要馬上回歸初始值。x_min 初始值要是很大的值，x_max 初始值是很小的值。

4.   要考慮座標有可能落在對稱線上，for 迴圈掃過所有點，某一點自己也可以跟自己取中點來做判斷。

完整程式碼：

// This code is created by EdocZec on 2020/01/25
#include <bits/stdc++.h>
#define _for(i,a,b) for(int i=(a); i<(b); ++i)   // this has been verified and can work correctly.
#define _dbg(arr,n)  _for(i,0,n){cout << " [Debug] arr[" << i << "] = " << (arr)[i] << endl;} // this has been verified and can work correctly.

using namespace std;
struct point
{
    int x;
    int y;
};
point a[4000];
int main(int argc, char** argv)
{
    int case_num;
    scanf("%d", &case_num);
    _for(i,0,case_num)
    {
       int num, x_min = 10001, x_max = -10001, center = 0;
       scanf("%d", &num);
       _for(i,0,num)
       {
           scanf("%d %d", &a[i].x, &a[i].y);
           //cout << " [Debug] read a[i].x == " << a[i].x << endl;
           if((a[i].x) > x_max)    x_max = a[i].x;
           if((a[i].x) < x_min)    x_min = a[i].x;
           a[i].x = a[i].x;
           a[i].y = a[i].y;
       }
       
       // cout << " [Debug] x_max == " << x_max << endl;
       // cout << " [Debug] x_min == " << x_min << endl;
       center = (x_max + x_min);
       // cout << " [Debug] center == " << center << endl;
       x_min = 10001;  x_max = -10001;
       // judge
       bool hit = false;
       _for(i,0,num)
       {
           hit = false;
           _for(j,0,num)
           {
               // cout << " [Debug] a[i].x == " << a[i].x << endl;
               // cout << " [Debug] a[j].x == " << a[j].x << endl;
               if(  ((a[i].x + a[j].x) == center  ) && 
                   (a[i].y == a[j].y)  )
               {
                   hit = true;
               }
           }
           if(!hit)
           {
               // cout << " [Debug] not hit." << endl;
               // cout << " [Debug] a[i].x == " << a[i].x << endl;
               cout << "NO" << endl;
               break;
           }
       }
       if(hit) cout << "YES" << endl;

    } // for each case.
    return 0;
}
// above ok, updated on 2020/01/25