1595 - Symmetry

The figure shownon the left is left-right symmetric as it is possible to foldthe 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 notleft-right symmetric as it is impossible to find such a vertical line.

Write a programthat determines whether a figure, drawn with dots, is left-right symmetric ornot. The dots are all distinct.

Input 

The input consistsof T test cases. The number of test cases T isgiven in the first line of the input file. The first line of each test casecontains an integer N , where N ( 1N1, 000) is the number ofdots in a figure. Each of the following N lines containsthe x-coordinate and y-coordinate of a dot. Both x-coordinatesand y-coordinates are integers between -10,000 and 10,000, bothinclusive.

Output 

Print exactly oneline for each test case. The line should contain `YES' if the figure isleft-right symmetric. and `NO', otherwise.

The followingshows 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

代码：

#include<iostream>

#include<vector>

#include<algorithm>

#include<map>

#include<cstring>

using namespacestd;

 

doubledot[1100][2];

 

int main()

{

    int test;

    cin>>test;

    while(test--)

    {

        vector<double> vec;

        map<pair<double,double>,int>  map1;

        memset(dot,0,sizeof(dot));

        int num;

        cin>>num;

        for(int i=0;i<num;i++)

        {

           cin>>dot[i][0]>>dot[i][1];

            vec.push_back(dot[i][0]);

            map1[make_pair(dot[i][0],dot[i][1])]=i;

        }

        sort(vec.begin(),vec.end());

        double x=(vec[0]+vec[vec.size()-1])/2;

        bool flag=true;

        for(int i=0;i<num;i++)

        {

           if(map1.count(make_pair(2*x-dot[i][0],dot[i][1]))==0)

            {

                cout<<"NO\n";

                flag=false;

                break;

            }

        }

        if(flag)

        {

            cout<<"YES\n";

        }

    }

    return 0;

}

由于规定了对称轴是垂直于X轴的，所以先求出X坐标，然后挨个判断。