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}](http://www.bkjia.com/uploads/allimg/150124/0914162R0-0.jpg)
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 ( 1N
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 <set>
#include <string>
using namespace std;
typedef pair<int,int> point;//定义类型;(不能够用预处理)
{
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)
}
/*
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;