hdu 3902 Swordsman 判断多边形是否轴对称以及对称轴个数

Problem Description

Mr. AC is a swordsman. His dream is to be the best swordsman in the world. To achieve his goal, he practices every day. There are many ways to practice, but Mr. AC likes “Perfect Cut” very much. The “Perfect Cut” can be described as the following two steps:
1.  Put a piece of wood block on the desk, and then suddenly wave the sword, cutting the block into two pieces.
2.  Without any motion, the two pieces must be absolutely axial symmetry.
According to the step two, when the board is an axial symmetry figure, Mr. AC has a chance to achieve the “Perfect Cut”. Now give you a board, and you should tell if Mr. AC has a chance to complete the “Perfect Cut”. The board is a simple polygon.

 

Input

The input contains several cases. 
The first line of one case contains a single integer n (3 <= n <= 20000), the number of points. The next n lines indicate the points of the simple polygon, each line with two integers x, y (0 <= x, y <= 20000). The points would be given either clockwise or counterclockwise.

 

Output

For each case, output the answer in one line. If Mr. AC has the chance, print “YES”, otherwise "NO".

 

Sample Input

  
  
  

   
   
   3
0 0
2 0
0 1
4
0 0
0 1
1 1
1 0
  
  
  

 

Sample Output

  
  
  

   
   
   NO
YES
  
  
  

//就算对称轴的个数

#include<iostream>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<cmath>
#include<cstdio>
using namespace std;


const int maxn=20010;
int N,M,Tot,next[maxn];
int x[maxn],y[maxn];
long long ans;


struct Point
{
    long long ed,ag;
}P[2*maxn],Q[maxn];


bool operator==(Point A,Point B)
{
    return A.ed==B.ed&&A.ag==B.ag;
}


void scanInt(int &x)
{
    char ch;
    while(ch=getchar(),ch<'0'||ch>'9');
    x=ch-&