中南大学第十一届大学生程序设计竞赛-COJ1896-Symmetry

1896: Symmetry

Submit Page Summary Time Limit: 1 Sec Memory Limit: 128 Mb Submitted: 2 Solved: 2

Description

We call a figure made of points is left-right symmetric as it is possible to fold the sheet of paper along a vertical line and to cut the figure into two identical halves.For example, if a figure exists five points, which respectively are (-2,5),(0,0),(2,3),(4,0),(6,5). Then we can find a vertical line x = 2 to satisfy this condition. But in another figure which are (0,0),(2,0),(2,2),(4,2), we can not find a vertical line to make this figure left-right symmetric.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 <=N <= 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.

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

Hint

Source

中南大学第十一届大学生程序设计竞赛

题目大意：给定二维平面上的n个点，判断这n个点是否左右对称，不需要考虑旋转对称的情况。

解题思路：先确定图的对称轴，将点分别从左到右和从右到左排序，检查对应位置上的点是否对称即可。

考查内容：排序函数的使用

时间复杂度： O(nlogn)

题目难度：★★

#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
int x,y;
}a[1005],b[1005];

bool cmp1(node u,node v)
{
return u.x==v.x?u.y<v.y:u.x<v.x;
}

bool cmp2(node u,node v)
{
return u.x==v.x?u.y<v.y:u.x>v.x;
}

int main()
{
ios::sync_with_stdio(false);
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i].x>>a[i].y;
b[i].x=a[i].x;
b[i].y=a[i].y;
}
sort(a+1,a+n+1,cmp1);
sort(b+1,b+n+1,cmp2);
int sym=a[1].x+b[1].x;
bool flag=true;
for(int i=1;i<=n;i++)
{
if(a[i].x+b[i].x!=sym||a[i].y!=b[i].y)
{
flag=false;
break;
}
}
if(flag) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}