Symmetry（对称轴存在问题）

最新推荐文章于 2021-04-30 02:59:17 发布

转载最新推荐文章于 2021-04-30 02:59:17 发布 · 612 阅读

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原文链接：http://www.cnblogs.com/hfc-xx/p/4657072.html

本文探讨了如何通过编程判断一组点在坐标系中是否围绕某条垂直线左右对称。通过对点集进行数学运算，利用双精度浮点数的精度特性，实现了一种避免精度问题的解决方案。

Symmetry

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Description

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

题意；在坐标系中给出的点中，寻找一个竖直的对称轴，让这些点关于它对称，如果它存在，输出YES，否则输出NO

看了很多人的博客，大多数此题用STL解决，了解它会更容易，但是我不太会，所以还是介绍的是一种自己好理解的避免出现精度问题的方法（坐标*2）

备注：

double类型进行运算会有精度误差.
比如
double a=1,b=1;
a-b可能不等于0，而是等于一个接近0的小数。所以我们认为只要a-b小于一个足够小的数(eps)的话，那么就可以认为a=b。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double eps=1e-5;//1e-5:浮点数,在计算机中这么表示,在数学中是科学计数法.1e-5的意思就是1乘以10的负5次幂。就是0.000001

int x[1010],y[1010];
int main()
{
　　int T;
　　cin>>T;
　　while(T--)
{
　　int n,i,j;;
　　cin>>n;

　　double sum=0;

for(i=1; i<=n; i++)
{
　　cin>>x[i]>>y[i];
　　sum+=x[i];
}
　　sum/=n;

for(i=1; i<=n; i++)
{
　　for(j=1; j<=n; j++)
{
　　if(abs(2*sum-x[i]-x[j])<eps&&abs(y[i]-y[j])<eps) //这里相当于判断2*sum是否等于x[i]+y[j]且y[i]是否等于y[j]
break;
}
　　if(j>n) break;
}
　　if(i>n)
cout<<"YES"<<endl;
　　else
cout<<"NO"<<endl;
}
}

转载于:https://www.cnblogs.com/hfc-xx/p/4657072.html