1595 - 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.

\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 

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

题解：刘汝佳白书135页练习题5.6；本题能够考虑暴力求解，也能够用set的查找来进行；


code：

#include <iostream>
#include <set>
#include <string>
using namespace std;
typedef pair<int,int> point;//定义类型；（不能够用预处理）

set<point> se;

int main()
{
    int cas;
    int n;
    int x,y;
    cin>>cas;
    while(cas--)
    {
        se.clear();
        cin>>n;
        int sum=0;
        for(int i=0; i<n; i++)
        {
            cin>>x>>y;
            sum+=x;//将全部横坐标加和；
            se.insert(point(x*n,y));//（备注1）
        }
        /*
        set<point> ::iterator it;
        for(it=se.begin(); it!=se.end(); ++it)
        {
            point p=*it;
            cout<<p.first<<"  "<<p.second<<endl;
        }
        */
        set<point> ::iterator it;
        bool flag = true;
        for(it=se.begin(); it!=se.end(); ++it)
        {
            point p=*it;
            if(se.find(point(2*sum-p.first,p.second))==se.end())
            {
                flag=false;
                break;
            }
        }
        flag?

cout<<"YES"<<endl: cout<<"NO"<<endl;
    }
    return 0;
}


备注1：

x*n的意思，就是存储的时候将横坐标扩大n倍，由于对称轴肯定是竖线，所以如果对称的话全部的横坐标加和再与n想除得出的应该就是答案的x坐标；之所以不是将sum/n。是由于考虑到精度的问题。除的话会出现double类型；换而言之，这个存储方式就是将全部横坐标扩大了n倍；