Symmetry 解题心得

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Description

 

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 10

5 10 

6 14

 

Sample Output 

 

YES 

NO 

YES


分析：我的思路是输入的同时，记下某一行的x的最大值和最小值，通过这一行的对称轴，再默认已它为所有行的对称轴去判断是否适用其他行，没有冲突即找到对称轴。


我的代码

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<map>

using namespace std;





struct point

{

    double x, y;



    point(int x1 = 0, int y1 = 0) :x(x1), y(y1)

    {}



    void mirror()

    {

        x = -x;

    }



    bool operator <(const point &b)const

    {

        if (x == b.x){

            return  (y< b.y);

        }

        else{

            return x < b.x;

        }

    }



}p[1010];





map<point, bool> m;



bool findmirror(point & p1, double x)

{

    point tempp;

    tempp.x = 2 * x - p1.x;

    tempp.y = p1.y;

    //auto result =m.find(tempp);

    if (m[tempp]){

        return true;

    }

    return false;

}



int Max = -15000;

int Min = 15000;





int main()

{

    int t;        

    cin >> t;

    while (t--){

        m.clear();

        bool flag = true;

        Max = -15000;

        Min = 15000;

        int n;

        cin >> n;

        

        //输入

        for (int i = 0; i < n; i++){

            cin >> p[i].x >> p[i].y;

            m[p[i]] = true;

            if (p[i].y == p[0].y){

                if (Max < p[i].x)                Max = p[i].x;

                if (Min>p[i].x)                Min = p[i].x;

            }

        }



    //判断

            double mir = (double)(Max + Min) / 2;

            for (int i = 0; i < n; i++){

                if (  !findmirror(p[i], mir)      ){

                    flag = false;

                    break;

                }

            }

            if (flag == true)

                puts("YES");

            else

                puts("NO");

        }

        







    return 0;

}

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