SPOJ 1812. Longest Common Substring II

1812. Longest Common Substring II

Problem 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:
alsdfkjfjkdsal
fdjskalajfkdsla
aaaajfaaaa

Output:
2

第一个串建SAM,剩下的串在上面跑....记录下每个节点的最小匹配长度(初始值是 len 即能表示的最长后缀),

从后向前更新fa节点的LCS....

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define ll long long
#define prt(k) cout<<#k"="<<k<<" ";

/**Suffix Automaton*/
const int Char=26,N=1000022;
struct SAM_Node
{
    SAM_Node *fa,*ch[Char];
    int len,id,pos;
    SAM_Node() {}
    SAM_Node(int _len)
    {
        fa=0; len=_len;
        memset(ch,0,sizeof ch);
    }
}pool[N],*root,*last,*tail;   ///tail
int SAM_size;
#define node SAM_Node
SAM_Node *newnode(int len)
{
    pool[SAM_size]=node(len);
    pool[SAM_size].id=SAM_size;
    return &pool[SAM_size++];
}
node *newnode(node *p)
{
    pool[SAM_size]=*p;
    pool[SAM_size].id=SAM_size;
    return &pool[SAM_size++];
}
void SAM_init()
{
    SAM_size=0;
    root=last=newnode(0);
    pool[0].pos=0;
}
void add(int x,int len)
{
    node *p=last,*np=newnode(p->len+1);
    np->pos=len; last=np;
    for(;p&&!p->ch[x];p=p->fa) p->ch[x]=np;
    if(!p) { np->fa=root;return ; }
    node *q=p->ch[x];
    if(q->len==p->len+1) { np->fa=q;return; }
    node *nq=newnode(q);
    nq->len=p->len+1;
    q->fa=nq; np->fa=nq;
    for(;p&&p->ch[x]==q;p=p->fa) p->ch[x]=nq;
}
void SAM_build(char *s)
{
    SAM_init();
    int len=strlen(s);
    for(int i=0;i<len;i++) add(s[i]-'a',i+1);
}
char a[N],b[N];
int c[N];
node *top[N];
int LCS[N];
void Max(int &a,int b) { a=max(a,b); }
void Min(int &a,int b) { a=min(a,b); }
int main()
{
    scanf("%s",a);
    SAM_build(a);
    for(int i=0;i<SAM_size;i++) c[pool[i].len]++;
    for(int i=1;i<=SAM_size;i++) c[i]+=c[i-1];
    for(int i=0;i<SAM_size;i++) top[--c[pool[i].len]]=&pool[i];
    int ans=0;
    while(scanf("%s",b)==1)
    {
        int ans=0,tmp=0;
        node *u=root;
        for(int i=0;b[i];i++)
        {
            int c=b[i]-'a';
            if(u->ch[c])
            {
                u=u->ch[c];
                tmp++;
            }
            else {
                while(u&&!u->ch[c]) u=u->fa;
                if(u)
                {
                    tmp=u->len+1;
                    u=u->ch[c];
                }
                else {
                    tmp=0;
                    u=root;
                }
            }
            Max(LCS[u->id],tmp);
        }
        for(int i=SAM_size-1;i>=0;i--)
        {
            Min(top[i]->len,LCS[top[i]->id]);
            if(top[i]->fa)
                Max(LCS[top[i]->fa->id],LCS[top[i]->id]);
            LCS[top[i]->id]=0;
        }
    }
    ans=0;
    for(int i=0;i<SAM_size;i++) Max(ans,pool[i].len);
    cout<<ans<<endl;
}