ABC335C - Loong Tracking-优快云博客

本文链接：https://blog.youkuaiyun.com/Brian_Pan_/article/details/135435949

文章描述了一种在蛇游戏中利用栈数据结构简化蛇头位置更新和查询的方法，避免了维护整个蛇身的复杂性，提高了查询和更新操作的时间复杂度至O(n+q)。

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A brute force solution to this problem would be modifying all $n$ body parts with each move. However, one might notice that we don’t have to maintain the entire snake body dynamically. Since each part will take the place of the previous part, the $i$ th part will always be $i - 1$ indices down from the snake head.

With this conclusion, we can simply update the snake head position in a stack-like data structure. In each new move, we append the new position of the snake head to the stack. During queries, the $i$ th body part would be the $i$ th element down the stack. This gives us $\mathcal O(n+q)$ complexity.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
const int Maxn=2e6+10;
int a[Maxn],b[Maxn];
int n,q,cnt;
inline int read()
{
	int s=0,w=1;
	char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')w-=1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
int main()
{
	// freopen("in.txt","r",stdin);
	n=read(),q=read(),cnt=n;
	for(int i=1;i<=n;++i)
	a[i]=n-i+1;
	while(q--)
	{
		int opt=read();
		if(opt==1)
		{
			// printf("mod\n");
			string s;
			cin>>s;
			int x=a[cnt],y=b[cnt];
			if(s[0]=='U')++y;
			if(s[0]=='D')--y;
			if(s[0]=='L')--x;
			if(s[0]=='R')++x;
			a[++cnt]=x;
			b[cnt]=y;
		}
		else
		{
			int p=read();
			printf("%d %d\n",a[cnt-p+1],b[cnt-p+1]);
		}
	}
	return 0;
}