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;

}