spoj1812 LCS2 - Longest Common Substring II

 

地址：http://www.spoj.com/problems/LCS2/

题面：

LCS2 - Longest Common Substring II

no tags 

 

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

Notice: new testcases added

思路：

　　对第一个穿构建sam，然后让后面的串在sam上走，同时记录每个串跑完后各个状态的所能匹配的最长长度，然后对所有串取个最小值，之后再取个最大值。

　　注意：如果状态p能够到达，那么状态fa[p]也必然能够到达，但可能在匹配时不会走到fa[p]状态，所以当p可达时，要更新fa[p]。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 struct SAM
 6 {
 7     static const int MAXN = 100001<<1;//大小为字符串长度两倍
 8     static const int LetterSize = 26;
 9 
10     int tot, last, ch[MAXN][LetterSize], fa[MAXN], len[MAXN];
11     int sum[MAXN], tp[MAXN]; //用于拓扑排序，tp为排序后的数组
12 
13     void init( void)
14     {
15         last = tot = 1;
16         len[1] = 0;
17         memset(ch,0,sizeof ch);
18         memset(fa,0,sizeof fa);
19     }
20 
21     void add( int x)
22     {
23         int p = last, np = last = ++tot;
24         len[np] = len[p] + 1;
25         while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
26         if( p == 0)
27             fa[np] = 1;
28         else
29         {
30             int q = ch[p][x];
31             if( len[q] == len[p] + 1)
32                 fa[np] = q;
33             else
34             {
35                 int nq = ++tot;
36                 memcpy( ch[nq], ch[q], sizeof ch[q]);
37                 len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
38                 while( p && ch[p][x] == q)  ch[p][x] = nq, p = fa[p];
39             }
40         }
41     }
42 
43     void toposort( void)
44     {
45         for(int i = 1; i <= len[last]; i++)   sum[i] = 0;
46         for(int i = 1; i <= tot; i++)   sum[len[i]]++;
47         for(int i = 1; i <= len[last]; i++)   sum[i] += sum[i-1];
48         for(int i = 1; i <= tot; i++)   tp[sum[len[i]]--] = i;
49     }
50 } sam;
51 char ss[100004];
52 int ans,mx[100001<<1],mi[100001<<1];
53 
54 int main(void)
55 {
56     //freopen("in.acm","r",stdin);
57     sam.init();
58     scanf("%s",ss);
59     for(int i=0,len=strlen(ss);i<len;i++)  sam.add(ss[i]-'a');
60     for(int i=1;i<=sam.tot;i++) mi[i]=1e6;
61     sam.toposort();
62     while(~scanf("%s",ss))
63     {
64         for(int i=0,len=strlen(ss),p=1,cnt=0;i<len;i++)
65         {
66             int c=ss[i]-'a';
67             if(sam.ch[p][c])
68                 p=sam.ch[p][c],cnt++;
69             else
70             {
71                 while(p&&!sam.ch[p][c]) p=sam.fa[p];
72                 if(!p)
73                     p=1,cnt=0;
74                 else
75                     cnt=sam.len[p]+1,p=sam.ch[p][c];
76             }
77             mx[p]=max(mx[p],cnt);
78         }
79         for(int i=sam.tot;i;i--)
80         {
81             int p=sam.tp[i];
82             mi[p]=min(mx[p],mi[p]);
83             if(mx[p]&&sam.fa[p]) mx[sam.fa[p]]=sam.len[sam.fa[p]];
84             mx[p]=0;
85         }
86     }
87     for(int i=1;i<=sam.tot;i++) ans=max(ans,mi[i]);
88     printf("%d\n",ans);
89     return 0;
90 }

 

posted @ 2017-09-14 00:40  weeping  阅读(192)  评论(0)  编辑  收藏  举报