【水讨论】#53 A. Square Earth?

最新推荐文章于 2023-02-26 22:50:07 发布

原创最新推荐文章于 2023-02-26 22:50:07 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#水题 #codeforces

CF-A题终结者同时被 2 个专栏收录

96 篇文章

订阅专栏

CF-A Terminator

88 篇文章

订阅专栏

本文探讨了在正方形边上确定两点间最短路径的问题，提供了详细的解析思路及代码实现，适用于竞赛编程与算法研究。

Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side n. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0, 0), (n, 0), (0, n) and (n, n).

Input

The single line contains 5 space-separated integers: n, x1, y1, x2, y2 (1 ≤ n ≤ 1000, 0 ≤ x1, y1, x2, y2 ≤ n) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square.

Output

You must print on a single line the shortest distance between the points.

Sample test(s)

input
2 0 0 1 0

output
1

input
2 0 1 2 1

output
4

input
100 0 0 100 100

output
200

有一个n为边长的正方形边上有两个点，问只能走正方形边界的情况下最短路程是多长（顺时针A到B或者逆时针A到B咯~）。

好吧……分情况讨论好了……

我们把这个正方形从左下到右上分为两半，然后——

1) 两个点在同侧的话，就坐标(x+y)的差即可，记住abs取绝对值

2) 在不同侧的话，就两个方向看看哪个小咯~一侧为x1+y1+x2+y2,另一侧就直接正方形周长4n来减去前面那个就好。

Code：

#include <cstdio>
#include <memory>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
	int n,x1,y1,x2,y2;
	cin>>n>>x1>>y1>>x2>>y2;
	int len1=x1+y1,len2=x2+y2; 
	//if((x1-y1)*(x2-y2)>=0) dist=abs(x1+y1-x2-y2); 	// 两点在 y=x 同侧 
	//else dist=x1+y1+x2+y2;				// 两点在 y=x 异侧 
	int dist=abs( x1+y1-( (x1-y1)*(x2-y2)>=0 ? (x2+y2):(-x2-y2) ) );
	cout<<min(dist,4*n-dist)<<endl;
	return 0;
}