【BZOJ2780】[Spoj]8093 Sevenk Love Oimaster

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?"

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

题意：给你一堆文本串，每次询问一个串在多少个文本串中出现过。

题解：多串匹配要用到广义SAM。就是在每当开始加入一个串的时候，将last指针变回root。

那么这题怎么搞？我们需要知道SAM中的每个节点被多少个文本串所包含。记录sum[i]表示i被多少个文本串包含，vis[i]表示当前时刻，最后一个包含i的文本串是哪个。在建完SAM后，我们将所有串在SAM上再跑一边，将经过的点，以及它的parent树上的所有祖先都更新一遍（因为一个点被影响后它的所有parent也要被影响），如果某个点的vis=当前时间，则退出，否则更新sum和vis。

时间复杂度我不太会证，大概O(nsqrt(n))吧？不过这题也有O(nlogn)的做法，就是求出parent树的DFS序，每次询问相当于问一个点在parent树的子树中有多少个不同的文本串，也就转换成在DFS序上的一段区间中有多少个不同的文本串。这个显然是HH的项链啊，不过感觉好麻烦。（懒）

#include <cstdio>

#include <cstring>

#include <iostream>

using namespace std;

int n,m,tot;

int pre[400010],ch[400010][26],mx[400010],s[400010],vis[400010],lp[400010],rp[400010];

char str[800010];

void updata(int x,int y)

{

	for(;x&&vis[x]!=y;x=pre[x])	s[x]++,vis[x]=y;

}

int extend(int x,int y,int p)

{

	int np=++tot;

	mx[np]=mx[p]+1;

	for(;p&&!ch[p][x];p=pre[p])	ch[p][x]=np;

	if(!p)	pre[np]=1;

	else

	{

		int q=ch[p][x];

		if(mx[q]==mx[p]+1)	pre[np]=q;

		else

		{

			int nq=++tot;

			pre[nq]=pre[q],pre[np]=pre[q]=nq,mx[nq]=mx[p]+1;

			memcpy(ch[nq],ch[q],sizeof(ch[q]));

			for(;p&&ch[p][x]==q;p=pre[p])	ch[p][x]=nq;

		}

	}

	return np;

}

int main()

{

	scanf("%d%d",&n,&m);

	int i,j,a,b;

	tot=1;

	for(i=1;i<=n;i++)

	{

		lp[i]=rp[i-1];

		scanf("%s",str+lp[i]);

		rp[i]=strlen(str);

		for(b=1,j=lp[i];j<rp[i];j++)	b=extend(str[j]-'a',i,b);

	}

	for(i=1;i<=n;i++)

		for(a=1,j=lp[i];j<rp[i];j++)	a=ch[a][str[j]-'a'],updata(a,i);

	for(i=1;i<=m;i++)

	{

		scanf("%s",str);

		a=strlen(str);

		for(b=1,j=0;j<a;j++)

		{

			if(ch[b][str[j]-'a'])	b=ch[b][str[j]-'a'];

			else	break;

		}

		if(j==a)	printf("%d\n",s[b]);

		else	printf("0\n");

	}

	return 0;

}