SPOJ LCS2 1812. Longest Common Substring II

SPOJ Problem Set (classical) 1812. Longest Common Substring IIProblem code: LCS2

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:alsdfkjfjkdsalfdjskalajfkdslaaaaajfaaaaOutput:2

Notice: new testcases added

Added:	Bin Jin
Date:	2007-09-24
Time limit:	2 s
Source limit:	50000B
Memory limit:	256 MB
Cluster:	Pyramid (Intel Pentium III 733 MHz)
Ages:	All protocol T: C ++ 4.0.0-8

Create a SAM for the first string, and run the remaining string on it... record the minimum matching length of each node (the initial value is the longest suffix that can be expressed by len ),

Update the LCS of the fa node from the back to the front ....

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=110000;struct SAM_Node{    SAM_Node *fa,*next[26];    int len,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;}char str[maxn],other[maxn];int LCS[maxn*2],len,n;int c[maxn*2]; SAM_Node *top[maxn*2];int main(){    scanf("%s",str);    len=strlen(str);    SAM_init();    for(int i=0;i<len;i++)        SAM_add(str[i]-'a',i+1);    for(int i=0;i<SAM_size;i++)        c[SAM_node[i].len]++;    for(int i=1;i<=len;i++)        c[i]+=c[i-1];    for(int i=0;i<SAM_size;i++)        top[--c[SAM_node[i].len]]=&SAM_node[i];    while(scanf("%s",other)!=EOF)    {        n++;        int len2=strlen(other);        SAM_Node* now=SAM_root;        int temp=0;        for(int i=0;i<len2;i++)        {            int x=other[i]-'a';            if(now->next[x])            {                temp++;                now=now->next[x];            }            else            {                while(now&&now->next[x]==0)                    now=now->fa;                if(now)                {                    temp=now->len+1;                    now=now->next[x];                }                else                {                    temp=0;                    now=SAM_root;                }            }            if(temp>LCS[now->id])                LCS[now->id]=temp;        }        for(int i=SAM_size-1;i>=0;i--)        {            if(LCS[top[i]->id]<top[i]->len)                top[i]->len=LCS[top[i]->id];            if(top[i]->fa&&LCS[top[i]->fa->id]<LCS[top[i]->id])               LCS[top[i]->fa->id]=LCS[top[i]->id];            LCS[top[i]->id]=0;        }    }    int ans=0;    for(int i=1;i<SAM_size;i++)    {        if(ans<SAM_node[i].len)            ans=SAM_node[i].len;    }    printf("%d\n",ans);    return 0;}