bzoj 2780: [Spoj]8093 Sevenk Love Oimaster （广义后缀自动机）

2780: [Spoj]8093 Sevenk Love Oimaster

Time Limit: 1 Sec   Memory Limit: 128 MB
Submit: 738   Solved: 257
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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

HINT

Source

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题解：广义后缀自动机

题目大意：给出n个原串，再给出m个查询串。求每个查询串出现在了多少原串中。

建立后缀自动机，所谓广义就是将几个串建立在同一个自动机上，每次加入一个新串的时候，将last恢复到root，然后其他照常做就可以了。

如何统计每个状态出现在了多少原串中呢？我们对于每个状态维护两个值size,nxt。nxt存储的是上一个到达这个状态的串是哪个原串，如果与当前的原串相同就不更新。如果不相同就size++.根据后缀自动机的性质，当前点的值改变，parent链上的信息都需要更改，链上nxt的值应该是区间分布的，所以只要找到一个nxt等于当前的点，就可以不用再往前找了。

#include
#include
#include
#include
#include
#define N 200003
using namespace std;
int ch[N][30],fa[N],l[N],n,m,len;
int r[N],v[N],cnt,np,p,nq,q,last,root,nxt[N],now,size[N];
char s[N];
void extend(int x)
{
	int c=s[x]-'a';
	p=last; np=++cnt; last=np; 
	l[np]=l[p]+1;
	for (;p&&!ch[p][c];p=fa[p]) ch[p][c]=np;
	if (!p) fa[np]=root;
	else {
		q=ch[p][c];
		if (l[q]==l[p]+1) fa[np]=q;
		else {
			nq=++cnt; l[nq]=l[p]+1;
			memcpy(ch[nq],ch[q],sizeof ch[nq]); size[nq]=size[q]; nxt[nq]=nxt[q];
			fa[nq]=fa[q];
			fa[q]=fa[np]=nq;
			for (;ch[p][c]==q;p=fa[p]) ch[p][c]=nq;
		}
	}
	for (;np;np=fa[np]) 
	 if (nxt[np]!=now) {
	 	size[np]++;
	 	nxt[np]=now;
	 }
	 else break;
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("my.out","w",stdout);
	scanf("%d%d",&n,&m);
	root=++cnt;
	for(int i=1;i<=n;i++) {
		scanf("%s",s+1);
		last=root;
		len=strlen(s+1);
		now=i;
		for (int j=1;j<=len;j++) 
		 extend(j);
	}
	for (int i=1;i<=m;i++) {
		scanf("%s",s+1);
		len=strlen(s+1);
		p=root;
		for (int j=1;j<=len;j++)  p=ch[p][s[j]-'a'];
	    printf("%d\n",size[p]);
	}
}