UVA1595 Symmetry(暴力模拟)

最新推荐文章于 2021-04-12 19:04:07 发布

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本文介绍了一种通过编程判断一组点是否能构成左右对称图形的方法。通过对点进行排序并比较对应点的位置来验证图形的对称性。

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题意：

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 (1 ≤ N ≤ 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.

Sample Input：

-2 5

0 0

6 5

4 0

2 3

0 4

4 0

0 0

5 14

6 10

5 10

6 14

Sample Output：

YES

分析：
题意就是给你一些点，是否能组成一个左右对称的图形，所以读题一定要仔细啊，我最开始只看到对称图形，还迷茫了一下难道要判断左右对称、上下对称、斜对称，那不是要死人了，再仔细读了下题才确定是左右对称，弱读题的菜鸡真的伤不起。我最开始想的方法有点复杂，是找到最中间的轴，把点排序后分为两部分，然后左右散开扫一遍是不是对应的，但是这又得考虑到同一个x位置上可能有多个点，就涉及到y轴上的分布，成功把自己绕了进去，看了大佬的解题报告，真的妙啊，把点集升序和降序分别排一遍，顺序扫一遍看dot1[i].x+dot[2].x是否等于中轴的两倍，因为y轴是按照同一个顺序排列的，y轴只需要考察是否相等就可以了，秒啊。中轴的找法是最左边的点和最右边的点的横坐标相加除以二。

AC代码:

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1010;
int t, n, l, r, temp;

struct node{
	int x,y;
};

node dot1[maxn], dot2[maxn];

bool cmp1(node a, node b){
	if(a.x == b.x)
		return a.y > b.y;
	return a.x < b.x;
}

bool cmp2(node a, node b){
	if(a.x == b.x)
		return a.y > b.y;
	return a.x > b.x;
}

bool ava(){
	for(int i = 0; i < n; i++){
		if(dot1[i].x + dot2[i].x != temp) return false;
		else if(dot1[i].y != dot2[i].y) return false;
	}
	return true;
}

int main(){
	cin>>t;
	while(t--){
		cin>>n;
		int a, b;
		l = 10010;
		r = -10010;
		for(int i = 0; i < n; i++){
			cin>>a>>b;
			dot1[i].x = dot2[i].x = a;
			dot1[i].y = dot2[i].y = b;
			if(a < l)  l = a;
			if(a > r)  r = a;
		}
		sort(dot1, dot1+n, cmp1);
		sort(dot2, dot2+n, cmp2);
		
		temp = l + r;
		
		if(ava()) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}