Symmetry

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：
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"cstdio"
#include"cstring"
#include"vector"
#include"algorithm"
using namespace std;
const int maxn=10000+10;

vector<int>pile[maxn];
int book[2*maxn+10];
int top,flag;

void Init()
{
    int n;
    cin>>n;
    memset(pile,0,sizeof(pile));
    memset(book,0,sizeof(book));
    top=1;
    int x,y;
    for(int i=0;i<n;i++)
    {
        cin>>x>>y;
        y+=10000;
        if(!book[y]) book[y]=top++;
        pile[book[y]].push_back(x);
    }
}

void Work()
{
    sort(pile[1].begin(),pile[1].end());
    int mid=(pile[1][0]+pile[1][pile[1].size()-1])/2;
    flag=0;
    for(int k=1;k<top;k++)
    {
        sort(pile[k].begin(),pile[k].end());
        for(int j=0;j<pile[k].size();j++)
        {
            if(((pile[k][j]+pile[k][pile[k].size()-j-1])/2)!=mid)
            {
                flag=1;
                break;
            }
        }
        if(flag) break;
    }
}

void Print()
{
    cout<<(flag==1?"NO":"YES")<<endl;
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
     Init();
     Work();
     Print();
    }
    return 0;
}