CF235C Cyclical Quest(SAM)

/*
统计串的出现次数显然可以在自动机上匹配出来即可
但是每次都挨个匹配的话会时间爆炸

那么考虑我们把串复制一份， 然后一起在后缀自动机上跑， 当我们匹配长度大于该串长度的时候强行失配即可

可能会有旋转后相同的串所以开个数组判重 


*/

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long 
#define M 2000100 
#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;
}
char s[M];
int ch[M][26], fa[M], len[M], sz[M], tot[M], a[M], lst = 1, cnt = 1;
int vis[M];

void insert(int c)
{
	int p = ++cnt, f = lst;
	lst = p;
	len[p] = len[f] + 1;
	sz[p] = 1;
	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;
			memcpy(ch[nq], ch[q], sizeof(ch[q]));
			fa[nq] = fa[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()
{
	scanf("%s", s + 1);
	int l = strlen(s + 1);
	for(int i = 1; i <= l; i++) insert(s[i] - 'a');
	for(int i = 1; i <= cnt; i++) tot[len[i]]++;
	for(int i = 1; i <= cnt; i++) tot[i] += tot[i - 1];
	for(int i = 1; i <= cnt; i++) a[tot[len[i]]--] = i;
	for(int i = cnt; i >= 1; i--) sz[fa[a[i]]] += sz[a[i]];
	int m = read();
	for(int cor = 1; cor <= m; cor++)
	{
		scanf("%s", s + 1);
		l = strlen(s + 1);
		for(int i = 1; i <= l; i++) s[l + i] = s[i];
		int now = 1, lenn = 0;
		ll ans = 0;
		for(int i = 1; i <= l * 2; i++)
		{
			int c = s[i] - 'a';
			while(now && !ch[now][c]) now = fa[now], lenn = len[now];
			if(!now) now = 1, lenn = 0;
			else now = ch[now][c], lenn++;
			while(now && len[fa[now]] >= l) now = fa[now], lenn = len[now]; //只能统计那些刚刚超过的节点 
			if(lenn >= l && vis[now] != cor) vis[now] = cor, ans += sz[now]; 
		}
		cout << ans << "
";
	}
	return 0;
}