Symmetry

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：
给出一张图，图上有一些点，你要做的就是找到一条线可以将这些点对称分开，如果能找到，输出yes，找不到输出no


思路：
最重要的一句“The figure on the right is not left-right symmetric as it is impossible to find such a vertical line”，vertical说明只需要垂直的竖线，那就好办多l
首先随便找到处于同一行的对称两点，通过它们求出中点，那么这条线就找到了。
然后枚举判断每一行的对称点到这条线的距离是否相等，如果不相等，马上break掉，输出no。
然后考虑数据结构，因为处在同一行的点你不知道会有多少，所以使用不定长数组vector会比较方便
再想到如果仅仅就以y为vector数组的下标，后面的枚举会不好进行，所以只需要开一个数组记录一下就好了，y的值是什么并不重要。
还有就是为了找到同一行对称两点，应该对那一行先排一下序，然后就好找了


代码：

#include"iostream"
#include"algorithm"
#include"cstring"
#include"vector"
using namespace std;
const int maxn=10000+10;
int a[10000+10000+10];
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
       int x,y,mid,top,flag;
       cin>>n;
       top=1;                           //top不能为0，因为if(!a[y]) a[y]=top++;
       flag=0;
       memset(a,0,sizeof(a));       
       vector<int> pile[maxn];     //这里不应该放在别处，每次输入n后都应该有一个全新的vector，以免受到上一次输入的影响
       for(int i=0;i<n;i++)
       {
       cin>>x>>y;
       y+=10000;
       if(!a[y]) a[y]=top++;
       pile[a[y]].push_back(x);
       }
       sort(pile[1].begin(),pile[1].end());
       mid=(pile[1][pile[1].size()-1]+pile[1][0])/2;
       for(int j=1;j<top;j++)
       {
           sort(pile[j].begin(),pile[j].end());
           for(int k=0;k<pile[j].size();k++)
           if((pile[j][k]+pile[j][pile[j].size()-1-k])/2!=mid)
           flag=1;
          if(flag==1) break;
       }
       if(flag==1)
       {
           cout<<"NO"<<endl;
       }
       else
       cout<<"YES"<<endl;
    }
    return 0;
}