【bzoj2780】[Spoj]8093 Sevenk Love Oimaster 广义后缀自动机

最新推荐文章于 2018-03-31 07:49:21 发布

转载最新推荐文章于 2018-03-31 07:49:21 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/GXZlegend/p/7114767.html

本文介绍如何使用广义后缀自动机解决一类字符串匹配问题：给定多个字符串及若干查询串，求查询串作为子串出现在哪些给定字符串中。通过构建广义后缀自动机并利用特定节点信息，实现高效查询。

题目描述

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?"

输入

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

输出

For each question, output the answer in one line.

样例输入

3 3
abcabcabc
aaa
aafe
abc
a
ca

样例输出

1
3
1

题目大意

给出n个字符串和q个询问，每次询问给定的字符串在多少个字符串中出现过（为多少个字符串的子串）。

题解

这种单字符串为多字符串的子串的问题，当然要用广义后缀自动机

对于后缀自动机上的每个节点，我们再开两个数组vis和si，记录一下它最后一个属于的字符串是什么，以及它属于多少个字符串。

那么当插入一个字符之后，应该更新它的vis和si信息，如果vis不等于当前字符串，则将vis更新，并将si++。由于它的parent树上的祖先节点与它有着相同信息，应该一起更新，即不断向上寻找fa，直到不存在fa或已经被更新过为止。

同时，当插入过程中新建节点nq时，由于它是用来代替q的，所以应该把q的vis、si赋给nq。

大量实践证明，这样做的时间复杂度是$O(n)$的（其实我也不太会证）

最后对于每个询问串，在后缀自动机中沿next不断查找，最后节点的si即为答案。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 400010
using namespace std;
int next[N][26] , fa[N] , dis[N] , vis[N] , si[N] , last , tot = 1;
char str[N];
void insert(int c , int t)
{
	int p = last , np = last = ++tot;
	dis[np] = dis[p] + 1;
	while(p && !next[p][c]) next[p][c] = np , p = fa[p];
	if(!p) fa[np] = 1;
	else
	{
		int q = next[p][c];
		if(dis[q] == dis[p] + 1) fa[np] = q;
		else
		{
			int nq = ++tot;
			memcpy(next[nq] , next[q] , sizeof(next[q])) , si[nq] = si[q] , vis[nq] = vis[q] , dis[nq] = dis[p] + 1 , fa[nq] = fa[q] , fa[np] = fa[q] = nq;
			while(p && next[p][c] == q) next[p][c] = nq , p = fa[p];
		}
	}
	for(p = np ; p && vis[p] != t ; p = fa[p]) vis[p] = t , si[p] ++ ;
}
int main()
{
	int n , m , i , j , l;
	scanf("%d%d" , &n , &m);
	for(i = 1 ; i <= n ; i ++ )
	{
		scanf("%s" , str) , l = strlen(str);
		for(j = 0 , last = 1 ; j < l ; j ++ ) insert(str[j] - 'a' , i);
	}
	while(m -- )
	{
		scanf("%s" , str) , l = strlen(str);
		for(i = 0 , j = 1 ; i < l ; i ++ ) j = next[j][str[i] - 'a'];
		printf("%d\n" , si[j]);
	}
	return 0;
}

转载于:https://www.cnblogs.com/GXZlegend/p/7114767.html