UVA 1595 Symmetry(暴力)

最新推荐文章于 2020-02-18 15:19:21 发布

github.com/4gnosed

最新推荐文章于 2020-02-18 15:19:21 发布

阅读量304

点赞数

CC 4.0 BY-SA版权

分类专栏： Online Judge 算法入门经典文章标签： uva 点对称

本文链接：https://blog.youkuaiyun.com/gnosed/article/details/78165839

Online Judge 同时被 2 个专栏收录

36 篇文章

订阅专栏

算法入门经典

20 篇文章

订阅专栏

本文介绍了一种通过编程方式判断二维平面上由点构成的图形是否左右对称的方法。通过对点集进行分析，确定是否存在一条垂直对称轴使得图形能够沿此轴折叠并形成两个完全相同的图形部分。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The figure shown on the left is left-right symmetric as it is possible to fold the sheet of paper along a vertical line, drawn as a dashed line, and to cut the figure into two identical halves. The figure on the right is not left-right symmetric as it is impossible to find such a vertical line.

$\epsfbox{p3226.eps}$

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 $\le$ N $\le$ 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.

The following shows sample input and output for three test cases.

Sample Input

Sample Output

YES 
NO 
YES

思路：

若存在对称抽，则其x坐标必定是所有点中最左边和最右边这两点的x坐标之和的一半（中点）。如何判断是否存在对称抽，就是判断所有点是否都有关于此假定对称抽对称的对应点（暴力枚取）。为避免中点x坐标是小数，输入所有点时x坐标乘以2。

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    int t,n,a,midx,maxx,minx,x[1005],y[1005];
    cin>>t;
    while(t--)
    {
        maxx=minx=0;
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>a>>y[i];
            x[i]=a*2;
            if(x[i]>x[maxx])    maxx=i;
            if(x[i]<x[minx])    minx=i;
        }

        midx=(x[maxx]+x[minx])/2;

        int i=0;
        for(;i<n;i++)
        {
            int flag=0;
            for(int j=0;j<n;j++)
            {
                if(y[i]==y[j]&&(x[i]+x[j]==midx*2))
                {
                    flag=1; break;
                }
            }
            if(!flag)   break;
        }

        if(i>=n)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}