BZOJ2780:[SPOJ8093]Sevenk Love Oimaster(广义SAM)

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, sevenk felt angry and began to check oimaster's online talk with ChuYuXun.    

Oimaster talked with ChuYuXun 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

Solution

先把大串的广义$SAM$建出来，然后用$n$个大串在$SAM$上跑。每个点开一个$vis[i]$和$size[i]$，存这个点上一次被哪个大串访问，这个点一共被几个大串访问过。

同时每访问一个点，就要沿着这个点的$fa$指针往上暴跳，更新$vis$，同时$size+1$。直到跳到一个$vis$是当前大串的点就停止。

答案就是用询问串在$SAM$上跑，终点的$size$值。

还有一个$nlogn$的做法我不会QAQ

Code

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #define N (201000)
 using namespace std;

 int n,m,l[N],r[N];
 char s[N<<],t[N<<];

 struct SAM
 {
     int son[N][],fa[N],step[N],size[N],vis[N];
     int p,q,np,nq,last,cnt;
     SAM(){last=cnt=;}

     void Insert(int x)
     {
         p=last; np=last=++cnt; step[np]=step[p]+;
         while (p && !son[p][x]) son[p][x]=np, p=fa[p];
         if (!p) fa[np]=;
         else
         {
             q=son[p][x];
             if (step[q]==step[p]+) fa[np]=q;
             else
             {
                 nq=++cnt; step[nq]=step[p]+;
                 memcpy(son[nq],son[q],sizeof(son[q]));
                 fa[nq]=fa[q]; fa[q]=fa[np]=nq;
                 while (son[p][x]==q) son[p][x]=nq,p=fa[p];
             }
         }
     }
     void Calc()
     {
         for (int i=; i<=n; ++i)
         {
             int now=;
             for (int j=l[i]; j<r[i]; ++j)
             {
                 now=son[now][s[j]-'a'];
                 int t=now;
                 while (t && vis[t]!=i) vis[t]=i,++size[t],t=fa[t];
             }
         }
     }
     void Find(char s[])
     {
         int now=;
         for (int j=,l=strlen(s); j<l; ++j)
             now=son[now][s[j]-'a'];
         printf("%d\n",size[now]);
     }
 }SAM;

 int main()
 {
     scanf("%d%d",&n,&m);
     for (int i=; i<=n; ++i)
     {
         scanf("%s",s+r[i-]); int len=strlen(s+r[i-]);
         l[i]=r[i-], r[i]=l[i]+len;
     }
     for (int i=; i<=n; ++i,SAM.last=)
         for (int j=l[i]; j<r[i]; ++j)
             SAM.Insert(s[j]-'a');
     SAM.Calc();
     for (int i=; i<=m; ++i)
         scanf("%s",t),SAM.Find(t);
 }