luoguP1368 工艺(最小表示法 后缀自动机）

最小表示法就是直接扫过去

后缀自动机就是每次找最字典序最小儿子输出

• 最小表示法

/*
最小表示法裸题， 我好像学过来着？？ 怎么忘得这么干净

 

*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long 
#define M 300010 
#define mmp make_pair
using namespace std;
int read()
{
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
int s[M];

int work(int n)
{
	int i = 0, j = 1, k = 0;
	while(i < n && j < n && k < n)
	{
		int t = s[(i + k) % n] - s[(j + k) % n];
		if(t == 0)
		{
			k++;
		}
		else
		{
			if(t > 0) i += k + 1;
			else j += k + 1;
			if(i == j) j++;
			k = 0;
		}
	}
	return i < j ? i : j;
}

int main()
{
	int len = read();
	for(int i = 0; i < len; i++) s[i] = read();
	int x = work(len);
	int tmp = -1;
	for(int y = x; y != tmp; y = (y + 1) % len)
	{
		cout << s[y] << " ";
		tmp = x;
		
	}
	return 0;
}

• 后缀自动机

/*
最小表示法裸题， 我好像学过来着？？ 怎么忘得这么干净

 

*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<map>
#define ll long long 
#define M 300010 
#define mmp make_pair
using namespace std;
int read()
{
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
int s[M];
int len[4 * M], fa[4 * M], cnt = 1, lst = 1;
map<int, int> ch[4 * M];
void insert(int c)
{
	int p = ++cnt, f = lst;
	lst = p;
	while(f && !ch[f][c]) ch[f][c] = p, f = fa[f];
	if(!f)
	{
		fa[p] = 1;
	}
	else
	{
		int q = ch[f][c];
		if(len[q] == len[f] + 1)
		{
			fa[p] = q;
		}
		else
		{
			int nq = ++cnt;
			fa[nq] = fa[q];
			ch[nq] = ch[q];
			len[nq] = len[f] + 1;
			fa[p] = fa[q] = nq;
			while(f && ch[f][c] == q) ch[f][c] = nq, f = fa[f]; 
		}
	}
}

int main()
{
	int len = read();
	for(int i = 0; i < len; i++) s[i] = read(), insert(s[i]);
	for(int i = 0; i < len; i++) insert(s[i]);
	int now = 1;
	for(int i = 0; i < len; i++)
	{
		cout << ch[now].begin()->first << " ";
		now = ch[now].begin()->second;
	}
	return 0;
}