poj3415 Common Substrings（后缀自动机）

Posted 2021-12-01 zhangbuang

A substring of a string T is defined as:

 

T( i, k)= T_iT_i +1... T_i+k -1, 1≤ i≤ i+k-1≤| T|.

 

Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):

 

S = {( i, j, k) | k≥ K, A( i, k)= B( j, k)}.

 

You are to give the value of |S| for specific A, B and K.

Input

The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.

1 ≤ |A|, |B| ≤ 10⁵
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.

 

Output

For each case, output an integer |S|.

Sample Input

2
aababaa
abaabaa
1
xx
xx
0

Sample Output

22
5

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对第一个串建立自动机，然后让第二个串在上面跑，记录一下到每个状态时的匹配的长度，
一个状态是很多后缀串的集合，我们只需要当前状态中大于等于K 小于等于匹配长度的串
然后如果fa有长度也大于等于k的串的话，那也可以去
为了避免每次都要向上找fa，打一个lazy标记一波，最后在一遍更新就行了。

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 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define ri register int
 6 using namespace std;
 7 const int N=2e5+5;
 8 typedef long long ll;
 9 int k,n;
10 char s[N];
11 inline int calc(char c){return c>=‘a‘&&c<=‘z‘?c-‘a‘:c-‘A‘+26;}
12 struct SAM{
13     int tot,rt,last,len[N],link[N],son[N][60],siz[N],lz[N],cnt[N],rk[N];
14     inline void init(){
15         memset(len,0,sizeof(len)),memset(siz,0,sizeof(siz)),memset(rk,0,sizeof(rk)),memset(son,0,sizeof(son));
16         memset(link,0,sizeof(link)),memset(cnt,0,sizeof(cnt)),memset(lz,0,sizeof(lz)),tot=last=rt=1,len[0]=-1;
17     }
18     inline void expend(int x){
19         int p=last,np=++tot;
20         siz[last=np]=1,len[np]=len[p]+1;
21         while(p&&!son[p][x])son[p][x]=np,p=link[p];
22         if(!p){link[np]=rt;return;}
23         int q=son[p][x],nq;
24         if(len[q]==len[p]+1){link[np]=q;return;}
25         len[nq=++tot]=len[p]+1,memcpy(son[nq],son[q],sizeof(son[q])),link[nq]=link[q];
26         while(p&&son[p][x]==q)son[p][x]=nq,p=link[p];
27         link[q]=link[np]=nq;
28     }
29     inline void topsort(){
30         for(ri i=1;i<=tot;++i)++cnt[len[i]];
31         for(ri i=1;i<=last;++i)cnt[i]+=cnt[i-1];
32         for(ri i=1;i<=tot;++i)rk[cnt[len[i]]--]=i;
33         for(ri i=tot;i;--i)siz[link[rk[i]]]+=siz[rk[i]];
34     }
35     inline void query(){
36         topsort();
37         int p=1,nowlen=0;
38         ll ans=0;
39         for(ri i=1;i<=n;++i){
40             int x=calc(s[i]);
41             if(son[p][x])p=son[p][x],++nowlen;
42             else{
43                 while(p&&!son[p][x])p=link[p];
44                 if(!p)p=1,nowlen=0;
45                 else nowlen=len[p]+1,p=son[p][x];
46             }
47             if(nowlen>=k){
48                 ans+=(ll)(nowlen-max(k,len[link[p]]+1)+1)*siz[p];
49                 if(len[link[p]]>=k)++lz[link[p]];
50             }
51         }
52         for(ri i=tot;i;--i){
53             p=rk[i];
54             ans+=(ll)lz[p]*siz[p]*(len[p]-max(k,len[link[p]]+1)+1);
55             if(len[link[p]]>=k)lz[link[p]]+=lz[p];
56         }
57         cout<<ans<<‘\n‘;
58     }
59 }sam;
60 int main(){
61     while(scanf("%d",&k),k){
62         scanf("%s",s+1),n=strlen(s+1),sam.init();
63         for(ri i=1;i<=n;++i)sam.expend(calc(s[i]));
64         scanf("%s",s+1),n=strlen(s+1),sam.query();
65     }
66     return 0;
67 }