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

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");elseputs("NO");}return 0;
}

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转载于:https://www.cnblogs.com/shawn-ji/p/4656817.html