UVA - 1595 Symmetry-优快云博客

本文链接：https://blog.youkuaiyun.com/zero_zp/article/details/76033939

本文介绍了一个用于判断点集是否关于垂直轴对称的程序设计思路。通过计算所有点的x坐标平均值来确定对称轴的位置，并检查每一点是否都有对应的对称点存在。

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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.

$\epsfbox{p3226.eps}$

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 $\le$ N $\le$ 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

Sample Output

YES 
NO 
YES
题意：是否所有的点都关于竖线对称。
题解：所有x的和除以n得到就是竖线的横坐标，关键是看一个点的对称点是否存在。
#include<bits/stdc++.h>
using namespace std;
struct node
{
    double a,b;
    node(double x,double y)
    {
        a=x,b=y;
    }
    friend bool operator < (node a,node b)
    {
        if(a.a!=b.a)
            return a.a<b.a;
        return a.b<b.b;
    }
};
int main()
{
    int T,n;
    double x[10011],y[10011];
    cin>>T;
    while(T--&&scanf("%d",&n)!=EOF)
    {
        map<node,int> m;
        double sx=0,sy=0;
        for (int i=0; i<n; i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
            sx+=x[i],sy+=y[i];
            node w(x[i],y[i]);
            m[w]++;
        }
        sx/=n,sy/=n;
        int i;
        for (i=0; i<n; i++)
        {
            double dx = fabs(sx-x[i]);
            node w1(sx+dx,y[i]),w2(sx-dx,y[i]);
            if(m[w1]!=m[w2])
                break;
        }
        if(i==n)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
原题链接：https://vjudge.net/contest/169852#problem/F