BZOJ2434 [NOI2011] 阿狸的打字机 【树链剖分】【线段树】【fail树】【AC自动机】

题目分析：

画一下fail树，就会发现就是x的子树中属于y路径的，把y剖分一下，用线段树处理

$O(n*log^2 n)$。

代码：

 #include<bits/stdc++.h>
 using namespace std;

 const int maxn = ;

 string str;
 struct node{int ch[],fail,fa;}T[maxn];
 int star[maxn],ans[maxn],num = ,snum,m;
 int dfsin[maxn],dfsout[maxn],poss[maxn]; // failtree
 int Tnb[maxn],top[maxn],sz[maxn];
 vector <int> g[maxn];
 vector<pair<int,int> > Qy[maxn];

 void read(){
     ios::sync_with_stdio(false);
     cin.tie();
     cin >> str;
     int now = ;
     for(int i=;i<str.length();i++){
     if(str[i] == 'B') now = T[now].fa;
     else if(str[i] == 'P') star[++snum] = now;
     else{
         if(T[now].ch[str[i]-'a']) now = T[now].ch[str[i]-'a'];
         else{
         T[now].ch[str[i]-'a'] = ++num;
         T[num].fa = now;
         now = num;
         }
     }
     }
     snum = ;
 }

 void dfs(int now){
     dfsin[now] = dfsout[now] = ++snum;
     poss[snum] = now;
     for(int i=;i<g[now].size();i++){
     dfs(g[now][i]);
     dfsout[now] = dfsout[g[now][i]];
     }
 }

 queue<int> q;
 void BuildACAutomaton(){
     q.push();T[].fail = ;
     while(!q.empty()){
     int k = q.front();q.pop();
     for(int i=;i<;i++){
         if(T[k].ch[i] == ) continue;
         int ff = T[k].fail;
         while(ff !=  && T[ff].ch[i] == ) ff = T[ff].fail;
         if(T[ff].ch[i]==T[k].ch[i]||T[ff].ch[i]==)T[T[k].ch[i]].fail=;
         else T[T[k].ch[i]].fail = T[ff].ch[i];
         q.push(T[k].ch[i]);
         g[T[T[k].ch[i]].fail].push_back(T[k].ch[i]);
     }
     }
     snum = ; dfs(); snum = ;
 }

 int QT[maxn*];

 void Add(int now,int tl,int tr,int pos){
     if(tl == tr){QT[now]++;return;}
     int mid = (tl+tr)/;QT[now]++;
     if(mid >= pos) Add(now<<,tl,mid,pos);
     else Add(now<<|,mid+,tr,pos);
 }
 int Query(int now,int tl,int tr,int l,int r){
     if(tl >= l && tr <= r) return QT[now];
     if(tl > r || tr < l) return ;
     int mid = (tl+tr)/;
     return Query(now<<,tl,mid,l,r)+Query(now<<|,mid+,tr,l,r);
 }

 void dfsmiao(int now){
     for(int i=;i<;i++){
     if(T[now].ch[i]==)continue;
     dfsmiao(T[now].ch[i]);
     sz[now] += sz[T[now].ch[i]];
     }
     sz[now]++;
 }

 void dfsyeah(int now,int tp){
     int son = ;top[now] = tp;Tnb[now] = ++snum;
     for(int i=;i<;i++){
     if(sz[T[now].ch[i]] > sz[son]) son = T[now].ch[i];
     }
     if(son) dfsyeah(son,tp);
     for(int i=;i<;i++){
     if(T[now].ch[i] ==  || T[now].ch[i] == son) continue;
     dfsyeah(T[now].ch[i],T[now].ch[i]);
     }
 }

 void treechain(){
     dfsmiao();
     dfsyeah(,);
 }

 void readquery(){
     cin >> m;
     for(int i=;i<=m;i++){
     int x,y; cin >> x >> y;
     x = star[x];y = star[y];
     Qy[poss[dfsin[x]-]].push_back(make_pair(y,i));
     Qy[poss[dfsout[x]]].push_back(make_pair(y,i));
     }
 }

 void solve(int endpos,int now){
     if(ans[now]) ans[now] = -ans[now];
     while(endpos){
     ans[now] += Query(,,num,Tnb[top[endpos]],Tnb[endpos]);
     endpos = T[top[endpos]].fa;
     }
 }

 void work(){
     for(int i=;i<=num;i++){
     int now = poss[i];
     Add(,,num,Tnb[now]);
     for(int j=;j<Qy[now].size();j++){
         solve(Qy[now][j].first,Qy[now][j].second);
     }
     }
     for(int i=;i<=m;i++) cout<<ans[i]<<endl;
 }

 int main(){
     //freopen("7.in","r",stdin);
     read();
     treechain();
     BuildACAutomaton();
     readquery();
     work();
     return ;
 }