题目给出N个字符串，我们现在想知道的是第x个字符串在第y个字符串中出现的次数是多少次？
求每个字符在其余字符中出现次数，想到从AC自动机上走，其实可以看作是求它的后缀的前缀中有多少个是满足的，换言之，我们可以去fail树中的父亲结点中查询。
就譬如说对于aa和aabaa两个字符串，我们搭建字典树
并且对应上fail指针，我们可以知道，aa出现的次数为两次，分别是以整个串aabaa构成的前缀，以及以aa（aabaa的后缀）构成的前缀。
搭建对应的fail树，我们知道每个对应的权值只能是以它的子树上的贡献，所以我们求它的子树中的某个节点的贡献，一定是它对应子树（含自己），所产生的贡献。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, len, M, tot, rid[maxN];
char s[maxN];
struct node
{
int nex[26], val, fail, fa;
inline void clear() { memset(nex, 0, sizeof(nex)); val = fail = fa = 0; }
}a[maxN];
inline void Insert()
{
int root = 0;
for(int i=0, ch; i<len; i++)
{
if(s[i] == 'P')
{
rid[++N] = root;
continue;
}
if(s[i] == 'B')
{
root = a[root].fa;
continue;
}
ch = s[i] - 'a';
if(!a[root].nex[ch])
{
++tot;
a[tot].clear();
a[tot].fa = root;
a[root].nex[ch] = tot;
}
root = a[root].nex[ch];
}
}
int head[maxN], cnt;
struct Eddge
{
int nex, to;
Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN];
inline void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
inline void build_fail()
{
queue<int> Q; Q.push(0);
int tmp, son, p;
while(!Q.empty())
{
tmp = Q.front(); Q.pop();
for(int i=0; i<26; i++)
{
son = a[tmp].nex[i];
if(son)
{
if(!tmp) a[son].fail = 0;
else
{
p = a[tmp].fail;
while(p && !a[p].nex[i]) p = a[p].fail;
a[son].fail = a[p].nex[i];
}
Q.push(son);
addEddge(a[son].fail, son);
}
else a[tmp].nex[i] = a[a[tmp].fail].nex[i];
}
}
}
int trie[maxN] = {0};
inline void update(int x, int val) { while(x < maxN) { trie[x] += val; x += lowbit(x); } }
inline int query(int x) { int ss = 0; while(x) { ss += trie[x]; x -= lowbit(x); } return ss; }
inline int Range_Query(int x, int y) { return query(y) - query(x - 1); }
int dfn[maxN] = {0}, _Index, siz[maxN];
void dfs(int u)
{
dfn[u] = ++_Index; siz[dfn[u]] = 1;
for(int i=head[u], v; ~i; i=edge[i].nex) { v = edge[i].to; dfs(v); siz[dfn[u]] += siz[dfn[v]]; }
}
struct Question
{
int x, y, id;
Question(int a=0, int b=0, int c=0):x(a), y(b), id(c) {}
friend bool operator < (Question e1, Question e2) { return e1.y < e2.y; }
}q[maxN];
int now_ques_ID, ans[maxN];
void dfs_Trie()
{
int root = 0, id_N = 0;
for(int i=0, ch; i<len; i++)
{
if(s[i] == 'P')
{
id_N++;
while(q[now_ques_ID].y == id_N)
{
int x = rid[q[now_ques_ID].x];
ans[q[now_ques_ID].id] = Range_Query(dfn[x], dfn[x] + siz[dfn[x]] - 1);
now_ques_ID++;
}
continue;
}
if(s[i] == 'B')
{
update(dfn[root], -1);
root = a[root].fa;
continue;
}
ch = s[i] - 'a';
root = a[root].nex[ch];
update(dfn[root], 1);
}
}
inline void init()
{
N = tot = cnt = _Index = 0;
now_ques_ID = 1;
a[0].clear();
memset(head, -1, sizeof(head));
}
int main()
{
scanf("%s", s); len = (int)strlen(s);
init();
Insert();
build_fail();
dfs(0);
scanf("%d", &M); int x, y;
for(int i=1; i<=M; i++)
{
scanf("%d%d", &x, &y);
q[i] = Question(x, y, i);
}
sort(q + 1, q + M + 1);
dfs_Trie();
for(int i=1; i<=M; i++) printf("%d\n", ans[i]);
return 0;
}
/*
aaPbaaP
1
1 2
ans:2
*/