UVA 1595 Symmetry(暴力)

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                                            -2 5                                         0 0 6 5 4 0 2 3 4 2 3 0 4 4 0 0 0 4 5 14 6 105 10 6 14

Sample Output 

YES NO YES

思路：

若存在对称抽，则其x坐标必定是所有点中最左边和最右边这两点的x坐标之和的一半（中点）。如何判断是否存在对称抽，就是判断所有点是否都有关于此假定对称抽对称的对应点（暴力枚取）。为避免中点x坐标是小数，输入所有点时x坐标乘以2。

#include <iostream>#include <cstdlib>using namespace std;int main(){    int t,n,a,midx,maxx,minx,x[1005],y[1005];    cin>>t;    while(t--)    {        maxx=minx=0;        cin>>n;        for(int i=0;i<n;i++)        {            cin>>a>>y[i];            x[i]=a*2;            if(x[i]>x[maxx])    maxx=i;            if(x[i]<x[minx])    minx=i;        }        midx=(x[maxx]+x[minx])/2;        int i=0;        for(;i<n;i++)        {            int flag=0;            for(int j=0;j<n;j++)            {                if(y[i]==y[j]&&(x[i]+x[j]==midx*2))                {                    flag=1; break;                }            }            if(!flag)   break;        }        if(i>=n)            cout<<"YES"<<endl;        else            cout<<"NO"<<endl;    }    return 0;}