BZOJ2780: [Spoj]8093 Sevenk Love Oimaster（广义后缀自动机，Parent树，Dfs序）

Description

Oimaster and sevenk love each other.

But recently,sevenk heard that a girl named ChuYuXun was dating with oimaster.As a woman's nature, s

evenk felt angry and began to check oimaster's online talk with ChuYuXun.    Oimaster talked with Ch

uYuXun n times, and each online talk actually is a string.Sevenk asks q questions like this,    "how

 many strings in oimaster's online talk contain this string as their substrings?"

有n个大串和m个询问，每次给出一个字符串s询问在多少个大串中出现过

Input

There are two integers in the first line, 

the number of strings n and the number of questions q.

And n lines follow, each of them is a string describing oimaster's online talk. 

And q lines follow, each of them is a question.

n<=10000, q<=60000 

the total length of n strings<=100000, 

the total length of q question strings<=360000

Output

For each question, output the answer in one line.

Sample Input

3 3
abcabcabc
aaa
aafe
abc
a
ca

Sample Output

1
3

1

解题思路：

利用Parent树的子节点都是父节点母串的性质，在大串的每个实节点打上标记，然后构建出这颗Parent树。

那么我们的问题就变成了在一个Fail节点子树内不同颜色个数。

像不像HH的项链？

把Dfs序构建出来，离线树状数组就好了。

代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 const int N=;
 const int M=;
 struct sant{
     int tranc[];
     int len;
     int pre;
 }s[N];
 struct pnt{
     int hd;
     int col;
     int ind;
     int oud;
 }p[N];
 struct ent{
     int twd;
     int lst;
 }e[N];
 struct qust{
     int l,r;
     int ans;
     int no;
 }q[N];
 int siz;
 int fin;
 int n,Q;
 int cnt;
 int dfn;
 char tmp[N];
 int line[N];
 int lst[N];
 int phr[N];
 int col[N];
 void Insert(int c,int whic)
 {
     int nwp,nwq,lsp,lsq;
     nwp=++siz;
     p[nwp].col=whic;
     s[nwp].len=s[fin].len+;
     for(lsp=fin;lsp&&!s[lsp].tranc[c];lsp=s[lsp].pre)
         s[lsp].tranc[c]=nwp;
     if(!lsp)
         s[nwp].pre=;
     else{
         lsq=s[lsp].tranc[c];
         if(s[lsq].len==s[lsp].len+)
             s[nwp].pre=lsq;
         else{
             nwq=++siz;
             s[nwq]=s[lsq];
             s[nwq].len=s[lsp].len+;
             s[nwp].pre=s[lsq].pre=nwq;
             while(s[lsp].tranc[c]==lsq)
             {
                 s[lsp].tranc[c]=nwq;
                 lsp=s[lsp].pre;
             }
         }
     }
     fin=nwp;
     return ;
 }
 void ade(int f,int t)
 {
     cnt++;
     e[cnt].twd=t;
     e[cnt].lst=p[f].hd;
     p[f].hd=cnt;
     return ;
 }
 void Dfs(int x)
 {
     p[x].ind=++dfn;
     phr[dfn]=x;
     col[dfn]=p[x].col;
     for(int i=p[x].hd;i;i=e[i].lst)
     {
         int to=e[i].twd;
         Dfs(to);
     }
     p[x].oud=dfn;
     return ;
 }
 bool cmp(qust x,qust y)
 {
     return x.r<y.r;
 }
 bool cmq(qust x,qust y)
 {
     return x.no<y.no;
 }
 int lowbit(int x)
 {
     return x&(-x);
 }
 void update(int pos,int x)
 {
     while(pos&&pos<=siz)
     {
         line[pos]+=x;
         pos+=lowbit(pos);
     }
     return ;
 }
 int query(int pos)
 {
     int ans=;
     while(pos)
     {
         ans+=line[pos];
         pos-=lowbit(pos);
     }
     return ans;
 }
 int main()
 {
     fin=++siz;
     scanf("%d%d",&n,&Q);
     for(int i=;i<=n;i++)
     {
         scanf("%s",tmp+);
         fin=;
         int len=strlen(tmp+);
         for(int j=;j<=len;j++)
             Insert(tmp[j]-'a',i);
     }
     for(int i=;i<=siz;i++)
         ade(s[i].pre,i);
     Dfs();
 
     for(int i=;i<=Q;i++)
     {
         scanf("%s",tmp+);
         int root=;
         int len=strlen(tmp+);
         for(int j=;j<=len;j++)
             root=s[root].tranc[tmp[j]-'a'];
         q[i].no=i;
         if(!root)
             continue;
         q[i].l=p[root].ind;
         q[i].r=p[root].oud;
     }
     std::sort(q+,q+Q+,cmp);
     int rr=;
     for(int i=;i<=Q;i++)
     {
         while(rr<=q[i].r)
         {
             if(!col[rr])
             {
                 rr++;
                 continue;
             }
             if(lst[col[rr]])
                 update(lst[col[rr]],-);
             update(rr,);
             lst[col[rr]]=rr;
             rr++;
         }
         rr--;
         if(!q[i].l)
             continue;
         q[i].ans=query(q[i].r)-query(q[i].l-);
     }
     std::sort(q+,q+Q+,cmq);
     for(int i=;i<=Q;i++)
         printf("%d\n",q[i].ans);
     return ;
 }