SPOJ Longest Common Substring

LCS - Longest Common Substring

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

Added by:	Bin Jin
Date:	2007-09-24
Time limit:	0.294s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel Pentium G860 3GHz)
Languages:	All except: C++ 5

 

 

 

 

 

 

 

 

解题：SAM求LCS

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 251000;
struct SAM{
    struct node{
        int son[26],f,len;
        void init(){
            f = -1;
            len = 0;
            memset(son,-1,sizeof son);
        }
    }sn[maxn<<1];
    int tot,last;
    void init(){
        tot = last = 0;
        sn[tot++].init();
    }
    int newnode(){
        sn[tot].init();
        return tot++;
    }
    void extend(int c){
        int np = newnode(),p = last;
        sn[np].len = sn[p].len + 1;
        while(p != -1 && sn[p].son[c] == -1){
            sn[p].son[c] = np;
            p = sn[p].f;
        }
        if(p == -1) sn[np].f = 0;
        else{
            int q = sn[p].son[c];
            if(sn[p].len + 1 == sn[q].len) sn[np].f = q;
            else{
                int nq = newnode();
                sn[nq] = sn[q];
                sn[nq].len = sn[p].len + 1;
                sn[q].f = sn[np].f = nq;
                while(p != -1 && sn[p].son[c] == q){
                    sn[p].son[c] = nq;
                    p = sn[p].f;
                }
            }
        }
        last = np;
    }
}sam;
char sa[maxn],sb[maxn];
int main(){
    while(~scanf("%s%s",sa,sb)){
        sam.init();
        int len1 = strlen(sa);
        int len2 = strlen(sb);
        for(int i = 0; i < len1; ++i)
            sam.extend(sa[i]-'a');
        int p = 0,ret = 0,clen = 0;
        for(int i = 0; i < len2; ++i){
            int k = sb[i] - 'a';
            if(sam.sn[p].son[k] != -1){
                clen++;
                p = sam.sn[p].son[k];
            }else{
               while(p != -1 && sam.sn[p].son[k] == -1) p = sam.sn[p].f;
                if(p == -1) clen = p = 0;
                else{
                    clen = sam.sn[p].len + 1;
                    p = sam.sn[p].son[k];
                }
            }
            ret = max(ret,clen);
        }
        printf("%d
",ret);
    }
    return 0;
}