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

/* ***********************************************
Author        :kuangbin
Created Time  :2013-9-8 23:27:46
File Name     :F:2013ACM练习专题学习后缀自动机
ewSPOJ_LCS.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int CHAR = 26;
const int MAXN = 250010;
struct SAM_Node
{
    SAM_Node *fa, *next[CHAR];
    int len;
    int id, pos;
    SAM_Node(){}
    SAM_Node(int _len)
    {
        fa = 0;
        len = _len;
        memset(next,0,sizeof(next));
    }
};
SAM_Node SAM_node[MAXN*2], *SAM_root, *SAM_last;
int SAM_size;
SAM_Node *newSAM_Node(int len)
{
    SAM_node[SAM_size] = SAM_Node(len);
    SAM_node[SAM_size].id = SAM_size;
    return &SAM_node[SAM_size++];
}
SAM_Node *newSAM_Node(SAM_Node *p)
{
    SAM_node[SAM_size] = *p;
    SAM_node[SAM_size].id = SAM_size;
    return &SAM_node[SAM_size++];
}
void SAM_init()
{
    SAM_size = 0;
    SAM_root = SAM_last = newSAM_Node(0);
    SAM_node[0].pos = 0;
}
void SAM_add(int x,int len)
{
    SAM_Node *p = SAM_last, *np = newSAM_Node(p->len + 1);
    np->pos = len;
    SAM_last = np;
    for(; p && !p->next[x];p = p->fa)
        p->next[x] = np;
    if(!p)
    {
        np->fa = SAM_root;
        return;
    }
    SAM_Node *q = p->next[x];
    if(q->len == p->len + 1)
    {
        np->fa = q;
        return;
    }
    SAM_Node *nq = newSAM_Node(q);
    nq->len = p->len + 1;
    q->fa = nq;
    np->fa = nq;
    for(; p && p->next[x] == q; p = p->fa)
        p->next[x] = nq;
}
void SAM_build(char *s)
{
    SAM_init();
    int len = strlen(s);
    for(int i = 0;i < len;i++)
        SAM_add(s[i] - 'a', i+1);
}
char str1[MAXN], str2[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(scanf("%s%s",str1,str2) == 2)
    {
        SAM_build(str1);
        int len = strlen(str2);
        SAM_Node *tmp = SAM_root;
        int ans = 0;
        int t = 0;
        for(int i = 0;i < len;i++)
        {
            if(tmp->next[str2[i]-'a'])
            {
                tmp = tmp->next[str2[i]-'a'];
                t++;
            }
            else
            {
                while(tmp && !tmp->next[str2[i]-'a'])
                    tmp = tmp->fa;
                if(tmp == NULL)
                {
                    tmp = SAM_root;
                    t = 0;
                }
                else
                {
                    t = tmp->len + 1;
                    tmp = tmp->next[str2[i]-'a'];
                }
            }
            ans = max(ans,t);
        }
        printf("%d
",ans);
    }
    return 0;
}