Symmetry UVa 1595（水题）

题目描述：

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 ( 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.

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

题目大意：

在平面内给至多1000个点，找到一个垂直于x轴的直线，使得这些点沿直线左右对称

题目思路：

把给出的点按x大小排序，之后找到中间的点（奇数个点）或者中间两个点的中轴线（偶数个点）判断是否沿这条直线对称就可以了（注意用double）

#include<stdio.h>
#include<algorithm>
using namespace std;
struct point { int x, y; }a[1005];
int n;
int sgn(double x)
{
if (fabs(x) < 1e-8)return 0;
else if (x > 0)return 1;
return -1;
}
int cmp(point a, point b)
{
return a.x < b.x;
}
int ok(double m)
{
int j;
for (int i = 0; i < n; i++)
{
double x1 = 2 * m - a[i].x, y1 = a[i].y;
for (j = 0; j < n; j++)
if (!sgn(a[j].x -x1)&&!sgn(a[j].y-y1))break;
if (j == n)return 0;
}
return 1;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d%d", &a[i].x, &a[i].y);
sort(a, a + n, cmp);
double x1 = a[(n - 1) / 2].x,x2=x1;
if(n%2==0)x2 = a[(n - 1) / 2 + 1].x;
if (ok(x1) || !(n % 2) &&ok((x1+x2)/2))printf("YES\n");
else printf("NO\n");
}
}