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最新推荐文章于 2021-05-18 00:21:34 发布

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本文介绍了一种用于判断二维平面上一系列点是否构成左右对称图形的算法。通过计算所有点的对称轴并检查每一点是否存在其对称点来实现。使用了pair和set数据结构来存储和查找点坐标，确保了算法的高效运行。

首先计算出对称轴：(最左边的横坐标+最右边的横坐标）/ 2 因为存在对称轴的话肯定是以俩个端点的对称轴为目标

然后遍历所有点，看每个点是否有它所对称的那个点，这个用set查找，提前输入时将所有点都放入set集合中

如果都能找到即为YES否则NO；

其中用到了pair，同时将x y轴坐标存入set集合，存x坐标时，所有点都乘二，避免除二时出现小数。

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.

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 ( 1N1, 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

#include<iostream>
#include<set>
using namespace std;
const int maxn=10005;
typedef pair <int,int> point;
set<point> xy;
int x[maxn],y[maxn],zhou;
int ok(int i,int n)
{
	int j;
	for(j=0;j<n;j++)
	if(y[i]==y[j] && (x[i]+x[j])/2==zhou) return 1;
	return 0;
}
int main()
{
	int t,n,i,j;
	cin>>t;
	while(t--)
	{
		int minx=999999,maxx=-999999;
		cin>>n;
		for(i=0;i<n;i++)
		{
			cin>>x[i]>>y[i];
			x[i]*=2;
			xy.insert(point(x[i],y[i]));
			minx = min(minx,x[i]);
            maxx = max(maxx,x[i]);
		}
		zhou=(minx+maxx)/2;
		for(i=0;i<n;i++)
		{
			if(!ok(i,n))
			break;
		}
		if(i>=n)
		cout<<"yes"<<endl;
		else
		cout<<"no"<<endl;
	}
}

-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