题意

给你一些串，还有一些询问
问你第x个串在第y个串中出现了多少次

思路

对这些串建ac自动机
根据fail树的性质：若x节点是trie中root到t任意一个节点的fail树的祖先，那么x一定是y的子串
而x在y中出现的次数为以x为fail树中的根节点的子树中，有多少个节点是trie树中根节点到y的
首先对询问离线
由于这题是一个节点一个节点建的ac自动机，所以我们可以根据这个建立的路径来维护一个当前节点到根的路径
路径上的点权为1，其余为0
每次到达一个询问到的y，我们就查询此时x的fail树中的子树和，这个可以预处理出fail树的dfs序用树状数组维护

但是如果排除这题的特殊性，要通过遍历trie的方式来更新答案
因为在操作fail的过程中，trie改变了

else trie[now][i]=trie[fail[now]][i];

也就是说bfs构造fail指针的时候，有时候会让trie指向之前的节点，也就是(trie中)高度比较小的节点
所以dfs只需要这样写

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);
}

代码

版本一

#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|1using 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initvoid 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 3a
aa
aa*/

版本二

#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|1using 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initvoid 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 8a
aa
aa*/