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<<2],t[N<<1];

struct SAM
{
    int son[N][28],fa[N],step[N],size[N],vis[N];
    int p,q,np,nq,last,cnt;
    SAM(){last=cnt=1;}

    void Insert(int x)
    {
        p=last; np=last=++cnt; step[np]=step[p]+1;
        while (p && !son[p][x]) son[p][x]=np, p=fa[p];
        if (!p) fa[np]=1;
        else
        {
            q=son[p][x];
            if (step[q]==step[p]+1) fa[np]=q;
            else
            {
                nq=++cnt; step[nq]=step[p]+1;
                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=1; i<=n; ++i)
        {
            int now=1;
            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=1;
        for (int j=0,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=1; i<=n; ++i)
    {
        scanf("%s",s+r[i-1]); int len=strlen(s+r[i-1]);
        l[i]=r[i-1], r[i]=l[i]+len;
    }
    for (int i=1; i<=n; ++i,SAM.last=1)
        for (int j=l[i]; j<r[i]; ++j)
            SAM.Insert(s[j]-'a');
    SAM.Calc();
    for (int i=1; i<=m; ++i)
        scanf("%s",t),SAM.Find(t);
}