题目链接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=2806
(luogu) https://www.luogu.org/problemnew/show/P4022

题解:对“作文库”中的串建广义SAM。(感觉加个#拼在一起直接SAM也行啊，只是常数大了点，但是大家都写的广义SAM我也就跟着写广义SAM了233333)

询问时二分\(L\), 变成求最少几个位置不匹配。然后DP方程是\(dp[i]=\min(dp[i-1]+1,\min_{i-match[i]\le j\le i-L} dp[j])\). 其中\(match[j]\)表示以\(j\)结尾最长能匹配的子串长度，用后缀自动机求出。单调队列优化即可。

应该a[i]-=96我写成a[i]-=48; dp完了之后不更新ans; 二分分反……我是有多无脑

注意一个问题是，由于这里有\(mid\)的限制，只有\(\le i-mid\)的位置才能被放进单调队列, 不可以一上来就把\(0\)放进去!

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 1.1e6;
const int S = 2;
int son[(N<<1)+3][S+1];
int fa[(N<<1)+3];
int len[(N<<1)+3];
char a[N+3];
char b[N+3];
int match[N+3];
int dp[N+3];
int dq[N+3];
int n,m,siz,rtn,lstpos;

void initSAM()
{
	siz = rtn = lstpos = 1;
}

void insertchar(char ch)
{
	int p = lstpos,np; siz++; np = lstpos = siz; len[np] = len[p]+1;
	for(; p && son[p][ch]==0; p=fa[p]) {son[p][ch] = np;}
	if(!p) {fa[np] = rtn;}
	else
	{
		int q = son[p][ch];
		if(len[q]==len[p]+1) {fa[np] = q;}
		else
		{
			siz++; int nq = siz; len[nq] = len[p]+1;
			memcpy(son[nq],son[q],sizeof(son[q]));
			fa[nq] = fa[q]; fa[np] = fa[q] = nq;
			for(; p && son[p][ch]==q; p=fa[p]) {son[p][ch] = nq;}
		}
	}
}

int main()
{
	initSAM();
	scanf("%d%d",&n,&m);
	for(int i=1; i<=m; i++)
	{
		lstpos = rtn;
		scanf("%s",a+1); int lena = strlen(a+1);
		for(int j=1; j<=lena; j++) {insertchar(a[j]-48);}
	}
	for(int i=1; i<=n; i++)
	{
		scanf("%s",a+1); int lena = strlen(a+1);
		for(int j=1; j<=lena; j++) {a[j]-=48;}
		int u = rtn,cur = 0;
		for(int j=1; j<=lena; j++)
		{
			while(u && son[u][a[j]]==0) {u = fa[u]; cur = len[u];}
			if(son[u][a[j]]) {u = son[u][a[j]]; cur++;}
			else {u = rtn; cur = 0;}
			match[j] = cur;
		}
		int left = 0,right = lena,mid,ans;
		while(left<right)
		{
			mid = (left+right+1)>>1;
			dp[0] = 0; int head = 1,tail = 0;
			if(mid<=1) {tail++; dq[tail] = 0;}
			for(int j=1; j<=lena; j++)
			{
				dp[j] = dp[j-1]+1;
				while(head<=tail && dq[head]<j-match[j]) {dq[head] = 0; head++;}
				if(head<=tail)
				{
					dp[j] = min(dp[j],dp[dq[head]]);
				}
				if(j-mid+1>=0)
				{
					while(head<=tail && dp[dq[tail]]>=dp[j-mid+1]) {dq[tail] = 0; tail--;}
					tail++; dq[tail] = j-mid+1;
				}
			}
			ans = dp[lena];
			if(ans*10<=lena) {left = mid;}
			else {right = mid-1;}
		}
		printf("%d\n",left);
	}
	return 0;
}