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 = 102000;

string str;
struct node{int ch[26],fail,fa;}T[maxn];
int star[maxn],ans[maxn],num = 1,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(0);
    cin >> str;
    int now = 1;
    for(int i=0;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 = 0;
}

void dfs(int now){
    dfsin[now] = dfsout[now] = ++snum;
    poss[snum] = now;
    for(int i=0;i<g[now].size();i++){
    dfs(g[now][i]);
    dfsout[now] = dfsout[g[now][i]];
    }
}

queue<int> q;
void BuildACAutomaton(){
    q.push(1);T[1].fail = 1;
    while(!q.empty()){
    int k = q.front();q.pop();
    for(int i=0;i<26;i++){
        if(T[k].ch[i] == 0) continue;
        int ff = T[k].fail;
        while(ff != 1 && T[ff].ch[i] == 0) ff = T[ff].fail;
        if(T[ff].ch[i]==T[k].ch[i]||T[ff].ch[i]==0)T[T[k].ch[i]].fail=1;
        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 = 0; dfs(1); snum = 0;
}

int QT[maxn*4];

void Add(int now,int tl,int tr,int pos){
    if(tl == tr){QT[now]++;return;}
    int mid = (tl+tr)/2;QT[now]++;
    if(mid >= pos) Add(now<<1,tl,mid,pos);
    else Add(now<<1|1,mid+1,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 0;
    int mid = (tl+tr)/2;
    return Query(now<<1,tl,mid,l,r)+Query(now<<1|1,mid+1,tr,l,r);
}

void dfsmiao(int now){
    for(int i=0;i<26;i++){
    if(T[now].ch[i]==0)continue;
    dfsmiao(T[now].ch[i]);
    sz[now] += sz[T[now].ch[i]];
    }
    sz[now]++;
}

void dfsyeah(int now,int tp){
    int son = 0;top[now] = tp;Tnb[now] = ++snum;
    for(int i=0;i<26;i++){
    if(sz[T[now].ch[i]] > sz[son]) son = T[now].ch[i];
    }
    if(son) dfsyeah(son,tp);
    for(int i=0;i<26;i++){
    if(T[now].ch[i] == 0 || T[now].ch[i] == son) continue;
    dfsyeah(T[now].ch[i],T[now].ch[i]);
    }
}

void treechain(){
    dfsmiao(1);
    dfsyeah(1,1);
}

void readquery(){
    cin >> m;
    for(int i=1;i<=m;i++){
    int x,y; cin >> x >> y;
    x = star[x];y = star[y];
    Qy[poss[dfsin[x]-1]].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(1,1,num,Tnb[top[endpos]],Tnb[endpos]);
    endpos = T[top[endpos]].fa;
    }
}

void work(){
    for(int i=1;i<=num;i++){
    int now = poss[i];
    Add(1,1,num,Tnb[now]);
    for(int j=0;j<Qy[now].size();j++){
        solve(Qy[now][j].first,Qy[now][j].second);
    }
    }
    for(int i=1;i<=m;i++) cout<<ans[i]<<endl;
}

int main(){
    //freopen("7.in","r",stdin);
    read();
    treechain();
    BuildACAutomaton();
    readquery();
    work();
    return 0;
}