A string is finite sequence of characters over a non-empty finite set Σ.
In this problem, Σ is the set of lowercase letters.
Substring, also called factor, is a consecutive sequence of characters occurrences at least once in a string.
Now your task is simple, for two given strings, find the length of the longest common substring of them.
Here common substring means a substring of two or more strings.
Input
The input contains exactly two lines, each line consists of no more than 250000 lowercase letters, representing a string.
Output
The length of the longest common substring. If such string doesn't exist, print "0" instead.
Example
Input: alsdfkjfjkdsal fdjskalajfkdsla Output: 3
n四次方的暴力不解释 = =
然后还有n方的dp
然后就是sa和sam
sa的复杂度是nlogn
然后就是sam了，玄学的复杂，完全不知道如何证明，o（玄学)就对了
‘
1 #include<iostream>
2 #include<cstdio>
3 #include<queue>
4 #include<map>
5 #include<set>
6 #include<queue>
7 #include<algorithm>
8 #include<cmath>
9 #include<stack>
10 #include<cstring>
11 #define mem(a,b) memset(a,b,sizeof a)
12 #define sc(a) scanf("%d",&(a))
13 #define scc(a,b) scanf("%d %d",&(a),&(b))
14 #define ll long long
15 using namespace std;
16
17
18 int kd[1000000];
19
20 struct dd
21 {
22
23 int fa,len;
24 int tr[29];
25
26 }dian[1200000];
27 int cnt[1200000];
28
29 int last=1;int tot=1;
30 inline void add(int c)
31 {
32 int p=last; int np=last=++tot;
33 dian[np].len=dian[p].len+1;
34
35 for(;p&&!dian[p].tr[c];p=dian[p].fa)dian[p].tr[c]=np;
36 if(!p)dian[np].fa=1,cnt[1]++;
37 else
38 {
39 int q=dian[p].tr[c];
40 if(dian[q].len==dian[p].len+1)dian[np].fa=q,cnt[q]++;
41 else
42 {
43 int nq=++tot;
44 dian[nq]=dian[q];
45 dian[nq].len=dian[p].len+1;
46 dian[q].fa=dian[np].fa=nq;
47 cnt[nq]+=2;
48 for(;p&&dian[p].tr[c]==q;p=dian[p].fa)dian[p].tr[c]=nq;
49
50 }
51 }
52 }
53 int ans=0;
54 void top()
55 {
56 int now=1,t=0;
57 string s;cin>>s; int len = s.length();
58 for(int i=0;i<len;i++)
59 {
60 int so=s[i]-'a';
61 if(dian[now].tr[so])now=dian[now].tr[so],t++;
62 else
63 {
64 while(now&&!dian[now].tr[so])now=dian[now].fa;
65 if(!now)now=1,t=0;
66 else{
67
68 t=dian[now].len+1;
69 //这里先对t赋值，因为这里是顺着fa向上走，到达的状态一定是两个串
70 //的公共字串，然后多了s【i】，t+=1;
71 now=dian[now].tr[so];
72 }
73 }
74 ans=max(ans,t);
75 }
76 cout<<ans;
77 }
78
79 signed main()
80 {
81 string s; string t;
82 //ios::sync_with_stdio(0);
83 // cin.tie(0);
84
85 cin>>t;
86 s+=t; int n=s.length();
87 for(int i=0;i<n;i++)add(s[i]-'a');
88 top();
89
90
91 }