| Time Limit: 236MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
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 a bit harder, for some 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 at most 10 lines, each line consists of no more than 100000 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 aaaajfaaaa Output: 2
【题意】
求若干个串的最长公共子串。
【思路】
SAM+DP
先拿个串建个SAM，然后用后面的串匹配，SAM中每个结点维护两个信息：mx表示与当前串匹配到该结点的最大长度，mn表示与前几个串匹配到该结点的最小长度。最后ans取所有节点mn的最大值。
注意mn[p]初始化为l[p]
cxlove：出现次数向父亲传递，接收串数从儿子获取
【代码】
1 #include<cstdio>
2 #include<cstring>
3 using namespace std;
4
5 const int N = 5*1e5+10;
6
7 char s[N];
8 int sz,last,root,l[N],ch[N][26],fa[N],mn[N],mx[N];
9 int b[N],cnt[N];
10 void add(int x) {
11 int c=s[x]-‘a‘;
12 int p=last,np=++sz; last=np;
13 l[np]=mn[np]=x+1;
14 for(;p&&!ch[p][c];p=fa[p]) ch[p][c]=np;
15 if(!p) fa[np]=root;
16 else {
17 int q=ch[p][c];
18 if(l[p]+1==l[q]) fa[np]=q;
19 else {
20 int nq=++sz; l[nq]=mn[nq]=l[p]+1;
21 memcpy(ch[nq],ch[q],sizeof(ch[q]));
22 fa[nq]=fa[q];
23 fa[np]=fa[q]=nq;
24 for(;ch[p][c]==q;p=fa[p]) ch[p][c]=nq;
25 }
26 }
27 }
28
29 int main() {
30 root=last=++sz;
31 scanf("%s",s);
32 int len=strlen(s);
33 for(int i=0;i<len;i++) add(i);
34 for(int i=1;i<=sz;i++) cnt[l[i]]++;
35 for(int i=1;i<=len;i++) cnt[i]+=cnt[i-1];
36 for(int i=1;i<=sz;i++) b[cnt[l[i]]--]=i;
37 while(scanf("%s",s)==1) {
38 int p=root; len=0;
39 for(int i=0;s[i];i++) {
40 int c=s[i]-‘a‘;
41 if(ch[p][c]) { len++; p=ch[p][c]; }
42 else {
43 while(p&&!ch[p][c]) p=fa[p];
44 if(!p) { len=0; p=root; }
45 else { len=l[p]+1; p=ch[p][c]; }
46 }
47 if(len>mx[p]) mx[p]=len;
48 }
49 for(int i=sz;i;i--) {
50 p=b[i];
51 if(mx[p]<mn[p]) mn[p]=mx[p];
52 if(fa[p] && mx[fa[p]]<mx[p]) mx[fa[p]]=mx[p];
53 mx[p]=0;
54 }
55 }
56 int ans=0;
57 for(int i=1;i<=sz;i++)
58 if(mn[i]>ans) ans=mn[i];
59 printf("%d",ans);
60 return 0;
61 }