bzoj 2434 fail tree+dfs序

　　首先比较明显的是我们可以将字符串组建立ac自动机，那么对于询问s1字符串在s2字符串中出现的次数，就是在以s1结尾为根的fail tree中，子树有多少个节点是s2的节点，这样我们处理fail tree的dfs序，然后用BIT维护，但是如果只是在线处理询问的话会tle，因为每个询问需要将节点的每一位在BIT中都修改，那么我们就浪费了好多性质，因为对于好多字符串拥有较长的LCP，那么我们可以模拟建trie的过程在自动机上跑，每遇到一个添加的字符，就解决所有询问为某字符串在该字符串中出现的次数，这样就可以了。

/**************************************************************
    Problem: 2434
    User: BLADEVIL
    Language: C++
    Result: Accepted
    Time:852 ms
    Memory:39128 kb
****************************************************************/
 
//By BLADEVIL
#include <cstdio>
#include <cstring>
#define maxn 200010
 
using namespace std;
 
struct node{
    int cnt,last,left,right;
    node *child[30],*fail,*father;
    node(){
        cnt=last=left=right=0;
        fail=father=NULL;
        memset(child,0,sizeof child);
    }
}nodepool[maxn],*totnode,*root,*que[maxn],*adr[maxn],*other[maxn],*adrans[maxn];
char c[maxn];
int len,tot,l,sum;
int pre[maxn],bit[maxn];
int ll,preans[maxn],otherans[maxn],lastans[maxn],sizeans[maxn];
int ans[maxn];
 
void add(int x,int y){
    while (x<=sum){
        bit[x]+=y;
        x+=x&(-x);
    }
}
 
int ask(int x){
    int ans=0;
    while (x){
        ans+=bit[x];
        x-=x&(-x);
    }
    return ans;
}
 
void connect(node *x,node *y){
    pre[++l]=x->last;
    x->last=l;
    other[l]=y;
    //printf("%d %d\n",x,y);
}
 
void connectans(int x,int y,int z){
    preans[++l]=lastans[x];
    lastans[x]=l;
    otherans[l]=y;
    sizeans[l]=z;
}
 
void build_trie(){
    totnode=nodepool; root=totnode++;
    scanf("%s",&c); len=strlen(c);
    node *t=root;
    for (int i=0;i<len;i++){
        if (c[i]=='P') adrans[++tot]=t; else
        if (c[i]=='B') t=t->father; else {
            if (!t->child[c[i]-'a']) 
                t->child[c[i]-'a']=totnode++,t->child[c[i]-'a']->father=t;
            t=t->child[c[i]-'a'];
            adr[i]=t;
        };
        //printf("%d ",t);
    }
    //printf("\n");
    //for (int i=0;i<len;i++) printf("%d ",adr[i]);
    //for (node *i=nodepool;i!=totnode;i++) printf("%d %d\n",i,i->father);
}
 
void build_ac(){
    int h=0,t=1;
    que[1]=root; root->fail=root;
    for (int i=0;i<26;i++) if (!root->child[i]) root->child[i]=root;
    while (h<t){
        node *v=que[++h];
        for (int i=0;i<26;i++) if (v->child[i]&&v->child[i]!=root){
            que[++t]=v->child[i];
            node *u=v->fail;
            que[t]->fail=u->child[i]!=que[t]?u->child[i]:root;
        } else v->child[i]=v->fail->child[i];
    }
    //for (int i=1;i<=t;i++) printf("%d %d ",que[i],que[i]->fail); printf("\n");
    //for (int i=1;i<=tot;i++) printf("%d ",adr[i]); printf("\n");
}
 
void dfs(node *x,node *fa){
    //printf("%d %d %d\n",x,fa,sum);
    x->left=++sum;
    for (int p=x->last;p;p=pre[p]){
        if (other[p]==fa) continue;
        dfs(other[p],x);
    }
    x->right=sum;
}
 
/*void work(){
    for (node *i=nodepool;i!=totnode;i++) if (i!=root) connect(i->fail,i);
    dfs(root,NULL);
    //for (node *i=nodepool;i!=totnode;i++) printf("%d %d %d %d\n",i,i->left,i->right,i->father);
    int m;
    scanf("%d",&m);
    while (m--){
        int x,y;
        scanf("%d %d",&x,&y);
        for (node *i=adr[y];i!=root;i=i->father) add(i->left,1);//printf("%d ",i->left);
        //printf("%d %d",adr[x]->left,adr[x]->right);
        //printf("%d ",ask(adr[x]->right));
        printf("%d\n",ask(adr[x]->right)-ask(adr[x]->left-1));
        for (node *i=adr[y];i!=root;i=i->father) add(i->left,-1);
    }
}*/
 
void work(){
    for (node *i=nodepool;i!=totnode;i++) if (i!=root) connect(i->fail,i);
    dfs(root,NULL);
    int m;
    scanf("%d",&m);
    for (int i=1;i<=m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        connectans(y,x,i);
    }
    int stack[maxn],tot=0,ansy=0;
    memset(stack,0,sizeof stack);
    for (int i=0;i<len;i++){
        if (c[i]=='P'){
            ansy++;
            for (int p=lastans[ansy];p;p=preans[p]){
                ans[sizeans[p]]=ask(adrans[otherans[p]]->right)-ask(adrans[otherans[p]]->left-1);
            }
        } else
        if (c[i]=='B') {
            add(adr[stack[tot--]]->left,-1);
        } else {
            stack[++tot]=i;
            add(adr[i]->left,1);
        }
    }
    for (int i=1;i<=m;i++) printf("%d\n",ans[i]);
}
 
int main(){
    build_trie();
    build_ac();
    work();
    return 0;
}