【BZOJ2780】[Spoj]8093 Sevenk Love Oimaster 广义后缀自动机

【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;
}
posted @ 2017-06-28 09:57  CQzhangyu  阅读(686)  评论(2编辑  收藏  举报