SPOJ 1811. Longest Common Substring （LCS，两个字符串的最长公共子串，后缀自动机SAM）

1811. Longest Common Substring

Problem code: LCS

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

后缀自动机SAM的模板题。

重新搞一遍SAM

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-9-8 23:27:46
  4 File Name     :F:2013ACM练习专题学习后缀自动机
ewSPOJ_LCS.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 
 21 const int CHAR = 26;
 22 const int MAXN = 250010;
 23 struct SAM_Node
 24 {
 25     SAM_Node *fa, *next[CHAR];
 26     int len;
 27     int id, pos;
 28     SAM_Node(){}
 29     SAM_Node(int _len)
 30     {
 31         fa = 0;
 32         len = _len;
 33         memset(next,0,sizeof(next));
 34     }
 35 };
 36 SAM_Node SAM_node[MAXN*2], *SAM_root, *SAM_last;
 37 int SAM_size;
 38 SAM_Node *newSAM_Node(int len)
 39 {
 40     SAM_node[SAM_size] = SAM_Node(len);
 41     SAM_node[SAM_size].id = SAM_size;
 42     return &SAM_node[SAM_size++];
 43 }
 44 SAM_Node *newSAM_Node(SAM_Node *p)
 45 {
 46     SAM_node[SAM_size] = *p;
 47     SAM_node[SAM_size].id = SAM_size;
 48     return &SAM_node[SAM_size++];
 49 }
 50 void SAM_init()
 51 {
 52     SAM_size = 0;
 53     SAM_root = SAM_last = newSAM_Node(0);
 54     SAM_node[0].pos = 0;
 55 }
 56 void SAM_add(int x,int len)
 57 {
 58     SAM_Node *p = SAM_last, *np = newSAM_Node(p->len + 1);
 59     np->pos = len;
 60     SAM_last = np;
 61     for(; p && !p->next[x];p = p->fa)
 62         p->next[x] = np;
 63     if(!p)
 64     {
 65         np->fa = SAM_root;
 66         return;
 67     }
 68     SAM_Node *q = p->next[x];
 69     if(q->len == p->len + 1)
 70     {
 71         np->fa = q;
 72         return;
 73     }
 74     SAM_Node *nq = newSAM_Node(q);
 75     nq->len = p->len + 1;
 76     q->fa = nq;
 77     np->fa = nq;
 78     for(; p && p->next[x] == q; p = p->fa)
 79         p->next[x] = nq;
 80 }
 81 void SAM_build(char *s)
 82 {
 83     SAM_init();
 84     int len = strlen(s);
 85     for(int i = 0;i < len;i++)
 86         SAM_add(s[i] - 'a', i+1);
 87 }
 88 char str1[MAXN], str2[MAXN];
 89 int main()
 90 {
 91     //freopen("in.txt","r",stdin);
 92     //freopen("out.txt","w",stdout);
 93     while(scanf("%s%s",str1,str2) == 2)
 94     {
 95         SAM_build(str1);
 96         int len = strlen(str2);
 97         SAM_Node *tmp = SAM_root;
 98         int ans = 0;
 99         int t = 0;
100         for(int i = 0;i < len;i++)
101         {
102             if(tmp->next[str2[i]-'a'])
103             {
104                 tmp = tmp->next[str2[i]-'a'];
105                 t++;
106             }
107             else
108             {
109                 while(tmp && !tmp->next[str2[i]-'a'])
110                     tmp = tmp->fa;
111                 if(tmp == NULL)
112                 {
113                     tmp = SAM_root;
114                     t = 0;
115                 }
116                 else
117                 {
118                     t = tmp->len + 1;
119                     tmp = tmp->next[str2[i]-'a'];
120                 }
121             }
122             ans = max(ans,t);
123         }
124         printf("%d
",ans);
125     }
126     return 0;
127 }