B - Symmetry

最新推荐文章于 2021-08-24 11:23:06 发布

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本文介绍了一种通过排序和比较来判断平面上一系列点是否构成左右对称图形的方法。通过对点进行排序，并寻找潜在的对称轴，可以高效地解决这一几何问题。

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We call a figure made of points is left-right symmetric as it is possible to fold the sheet of paper along a vertical line and to cut the figure into two identical halves.For example, if a figure exists five points, which respectively are (-2,5),(0,0),(2,3),(4,0),(6,5). Then we can find a vertical line x = 2 to satisfy this condition. But in another figure which are (0,0),(2,0),(2,2),(4,2), we can not find a vertical line to make this figure left-right symmetric.Write a program that determines whether a figure, drawn with dots, is left-right symmetric or not. The dots are all distinct.

Input

The input consists of T test cases. The number of test cases T is given in the first line of the input file. The first line of each test case contains an integer N, where N (1 <=N <= 1, 000) is the number of dots in a figure. Each of the following N lines contains the x-coordinate and y-coordinate of a dot. Both x-coordinates and y-coordinates are integers between −10, 000 and 10, 000, both inclusive.

Output

Print exactly one line for each test case. The line should contain 'YES' if the figure is left-right symmetric,and 'NO', otherwise.

Sample Input

Sample Output

YES
NO
YES

题目大意：在一个平面上有n个点，判断这n个点是否对称。

刚开始我也没有思路，看了一下别人的代码才知道原来还要用到排序。

思路：先确定图的对称轴，然后将点从左到右和从右到左排序，检查对应位置上的点即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

const int N=1000+5;
struct node
{
   int x;
   int y;

}a[N],b[N];

bool cmp0(node w,node h)
{
    if(w.x==h.x)
        return w.y<h.y;
    else
        return w.x<h.x;
}

bool cmp1(node w,node h)
{
    if(w.x==h.x)
        return w.y<h.y;
    else
        return w.x>h.x;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
      int n,sum,flag=1;
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
      {
          cin>>a[i].x>>a[i].y;
          b[i].x=a[i].x;
          b[i].y=a[i].y;
      }
       sort(a+1,a+n+1,cmp0);
       sort(b+1,b+n+1,cmp1);
       sum=a[1].x+b[1].x;
       for(int i=1;i<=n;i++)
       {
           if(a[i].x+b[i].x!=sum||a[i].y!=b[i].y)
           {
               flag=0;
               break;
           }
       }
       if(flag)
        printf("YES\n");
       else
        printf("NO\n");
    }
    return 0;
}