The meaning of problems
Give you some string, and some asked
to ask you the x-th string appeared in the y-th string how many times
Thinking
These strings built ac automatic machine
according to the nature fail tree: if the x node is a trie root to t any node fail tree ancestors, then x must be the y substring
number and x occurs in y as In x for the sub-tree root node fail tree, how many nodes are trie tree root node y to the
first interrogation offline
Since this is a question of a node built ac automatic machine, so we can build this path to maintain a current path to the root node
point on the right path is 1, and 0
each time a query to reach y, we fail query case subtree in the tree and x, this can be a pre-fail dfs tree sequence with Fenwick tree maintenance
However, if you exclude the special nature of this question, the answer should be updated by way of traversing the trie
because in the course of the operation fail, trie changed
else trie[now][i]=trie[fail[now]][i];That bfs construction fail pointer when sometimes make trie node before the point, which is (trie in) height smaller nodes
so dfs only need to write
void dfs(int x){
//printf("==%d\n",x);
add(bg[x],1);
for(int i = 0; i < (int)Q[x].size(); i++){
int y = Q[x][i].fst;
ans[Q[x][i].sc]=sum(ed[y])-sum(bg[y]-1);
}
for(int i = 0; i < 26; i++){
int y = ac.trie[x][i];
if(!y||h[y]<=h[x])continue;
dfs(y);
}
add(bg[x],-1);
}Code
A version
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
using namespace std;
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 3e5+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
vector<int>v[maxn];
int id[maxn];
struct AC{
//局部变量没有默认0!
int trie[maxn][26];
int num[maxn];//单词出现次数
int fail[maxn],fa[maxn];
int vis[maxn];//ask函数用到
int tot,rt;
int ntot;
//多测可写个init
void init(){tot=0;mem(vis,0);mem(trie,0);rt=0;ntot=0;}
void add(char c){
if(c=='P'){
num[rt]++;
id[++ntot]=rt;
return;
}
if(c=='B'){
rt=fa[rt];
return;
}
int x = c-'a';
if(!trie[rt][x]){
trie[rt][x]=++tot;
fa[tot]=rt;
}
//fa[trie[rt][x]]=rt;
rt=trie[rt][x];
}
void build(){
queue<int>q;
for(int i = 0; i < 26; i++){
if(trie[0][i]){
//fail[trie[0][i]]=0;
//v[0].pb(trie[0][i]);
q.push(trie[0][i]);
}
}
while(!q.empty()){
int now = q.front();
q.pop();
for(int i = 0; i < 26; i++){
////让这个节点的失败指针指向(((他父亲节点)的失败指针所指向的那个节点)的下一个相同节点)
if(trie[now][i]){
fail[trie[now][i]]=trie[fail[now]][i];
//v[trie[fail[now]][i]].pb(trie[now][i]);
q.push(trie[now][i]);
}
else trie[now][i]=trie[fail[now]][i];
//否则就让当前节点的这个子节点指向当前节点fail指针的这个子节点
}
v[fail[now]].pb(now);
}
}
}ac;
char a[maxn];
int bg[maxn],ed[maxn];
int S[maxn],tot;
void gao(int x){
S[++tot]=x;
bg[x]=tot;
for(int i = 0; i < (int)v[x].size(); i++){
int y =v[x][i];
gao(y);
}
//S[++tot]=x;
ed[x]=tot;
}
int tree[maxn];
int lowbit(int x){return x&-x;}
void add(int x, int c){
for(int i = x; i <= tot; i+=lowbit(i)){tree[i]+=c;}
}
int sum(int x){
int ans =0;
for(int i = x; i; i-=lowbit(i))ans+=tree[i];
return ans;
}
vector<PI>Q[maxn];
int ans[maxn];
/*
void dfs(int x){
//printf("%d\n",x);
add(bg[x],1);
for(int i = 0; i < (int)Q[x].size(); i++){
int y = Q[x][i].fst;
ans[Q[x][i].sc]=sum(ed[y])-sum(bg[y]-1);
}
for(int i = 0; i < 26; i++){
int y = ac.trie[x][i];
if(!y||(y==ac.trie[ac.fail[x]][i]))continue;
dfs(y);
}
add(bg[x],-1);
}*/
void sv(){
int now = 0;
int n = strlen(a+1);
int world=0;
for(int i = 1; i <= n; i++){
if(a[i]=='B'){
add(bg[now],-1);
now=ac.fa[now];
continue;
}
else if(a[i]=='P'){
world++;
for(int j = 0; j < (int)Q[now].size(); j++){
int x = Q[now][j].fst;
int y = Q[now][j]. sc;
ans[y]=sum(ed[x])-sum(bg[x]-1);
}
continue;
}
else{
int x=a[i]-'a';
now=ac.trie[now][x];
add(bg[now],1);
}
}
}
int main(){
ac.init();
tot=0;
scanf("%s", a+1);
int n = strlen(a+1);
for(int i = 1; i <= n; i++){
ac.add(a[i]);
}
ac.build();
int m;
scanf("%d", &m);
for(int i = 1; i <= m; i++){
int x,y;
scanf("%d %d", &x, &y);
x=id[x];y=id[y];
Q[y].pb(make_pair(x,i));
}
gao(0);
//dfs(0);
sv();
for(int i = 1; i <= m; i++){
printf("%d\n",ans[i]);
}
return 0;
}
/*
aPaPBbP
3
1 2
1 3
2 3
a
aa
aa
*/Version two
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
using namespace std;
typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
const db eps = 1e-6;
const int mod = 1e9+7;
const int maxn = 3e5+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);
vector<int>v[maxn];
int id[maxn];
int h[maxn];
struct AC{
//局部变量没有默认0!
int trie[maxn][26];
int num[maxn];//单词出现次数
int fail[maxn],fa[maxn];
int vis[maxn];//ask函数用到
int tot,rt;
int ntot;
int H;
//多测可写个init
void init(){tot=0;mem(vis,0);mem(trie,0);rt=0;ntot=0;H=0;}
void add(char c){
if(c=='P'){
num[rt]++;
id[++ntot]=rt;
return;
}
if(c=='B'){
rt=fa[rt];H--;
return;
}
int x = c-'a';
if(!trie[rt][x]){
trie[rt][x]=++tot;
fa[tot]=rt;
}
//fa[trie[rt][x]]=rt;
rt=trie[rt][x];
h[rt]=++H;
}
void build(){
queue<int>q;
for(int i = 0; i < 26; i++){
if(trie[0][i]){
fail[trie[0][i]]=0;
//v[0].pb(trie[0][i]);
q.push(trie[0][i]);
}
}
while(!q.empty()){
int now = q.front();
q.pop();
for(int i = 0; i < 26; i++){
////让这个节点的失败指针指向(((他父亲节点)的失败指针所指向的那个节点)的下一个相同节点)
if(trie[now][i]){
fail[trie[now][i]]=trie[fail[now]][i];
//v[trie[fail[now]][i]].pb(trie[now][i]);
q.push(trie[now][i]);
}
else trie[now][i]=trie[fail[now]][i];
//否则就让当前节点的这个子节点指向当前节点fail指针的这个子节点
}
v[fail[now]].pb(now);
}
}
}ac;
char a[maxn];
int bg[maxn],ed[maxn];
int S[maxn],tot;
void gao(int x){
S[++tot]=x;
bg[x]=tot;
for(int i = 0; i < (int)v[x].size(); i++){
int y =v[x][i];
gao(y);
}
//S[++tot]=x;
ed[x]=tot;
}
int tree[maxn];
int lowbit(int x){return x&-x;}
void add(int x, int c){
for(int i = x; i <= tot; i+=lowbit(i)){tree[i]+=c;}
}
int sum(int x){
int ans =0;
for(int i = x; i; i-=lowbit(i))ans+=tree[i];
return ans;
}
vector<PI>Q[maxn];
int ans[maxn];
void dfs(int x){
//printf("==%d\n",x);
add(bg[x],1);
for(int i = 0; i < (int)Q[x].size(); i++){
int y = Q[x][i].fst;
ans[Q[x][i].sc]=sum(ed[y])-sum(bg[y]-1);
}
for(int i = 0; i < 26; i++){
int y = ac.trie[x][i];
if(!y||h[y]<=h[x])continue;
dfs(y);
}
add(bg[x],-1);
}
void sv(){
int now = 0;
int n = strlen(a+1);
int world=0;
for(int i = 1; i <= n; i++){
if(a[i]=='B'){
add(bg[now],-1);
now=ac.fa[now];
continue;
}
else if(a[i]=='P'){
world++;
for(int j = 0; j < (int)Q[now].size(); j++){
int x = Q[now][j].fst;
int y = Q[now][j]. sc;
ans[y]=sum(ed[x])-sum(bg[x]-1);
}
continue;
}
else{
int x=a[i]-'a';
now=ac.trie[now][x];
add(bg[now],1);
}
}
}
int main(){
ac.init();
tot=0;
scanf("%s", a+1);
int n = strlen(a+1);
for(int i = 1; i <= n; i++){
ac.add(a[i]);
}
ac.build();
int m;
scanf("%d", &m);
for(int i = 1; i <= m; i++){
int x,y;
scanf("%d %d", &x, &y);
x=id[x];y=id[y];
Q[y].pb(make_pair(x,i));
}
gao(0);
dfs(0);
//sv();
for(int i = 1; i <= m; i++){
printf("%d\n",ans[i]);
}
return 0;
}
/*
asPdPasPddPhBdPhPnaPasP
8
1 5
2 6
3 8
4 3
7 7
2 5
6 8
1 8
a
aa
aa
*/