Symmetry（对称轴存在问题）

                    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.

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 ( 1N1, 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 

3                                            
5                                            
-2 5                                         
0 0 
6 5 
4 0 
2 3 
4 
2 3 
0 4 
4 0 
0 0 
4 
5 14 
6 10
5 10 
6 14

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;
}
}