spoj LCS2（多个串的最长公共子序列，后缀自动机）

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
Notice: new testcases added

多个串的最长公共子序列
和两个串的套路一样，每次记录能匹配到的点，能匹配到的点是一些串的集合，你能到这个点证明这个公共串必定是这个点的串集合里的串之一，那么问题来了。所以每个点记录一下Maxi，为这个点可以在每个串和A串匹配时可以接收的最长公共子串，同时还要往上更新fail 因为当前节点要是可以接受这个子串，那么fail结点必然也可以结束，而且接受的最大长度为max(Maxi[fail[p]],Maxi[p]),但是要和能到当前点的最大长度取min，统计每个串的时候都要对这个最大的取min，这样才是共有的最长子串。最后统计一遍答案就好

#include <bits/stdc++.h>using namespace std;const int maxn = 500000+5;int last,tail,in[maxn],Min[maxn];int Max[maxn],cnt[maxn],vis[maxn];int nxt[maxn][26],fail[maxn];char sa[maxn],sb[maxn];int Maxi[maxn],Mini[maxn];inline void build(char *s){    while(*s)    {        int p=last,t=++tail,c=*s++-'a';        Max[t]=Max[p]+1;        Mini[t]=Max[t];        Maxi[t]=0;        while(p&&!nxt[p][c])            nxt[p][c]=t,p=fail[p];        if(p)        {            int q=nxt[p][c];            if(Max[q]==Max[p]+1)                fail[t]=q,Min[t]=Max[q]+1;            else            {                int k=++tail;                fail[k]=fail[q];                fail[t]=fail[q]=k;                Max[k]=Max[p]+1;                Mini[k]=Max[k];                Maxi[k]=0;                memcpy(nxt[k],nxt[q],sizeof(nxt[q]));                while(p&&nxt[p][c]==q)                    nxt[p][c]=k,p=fail[p];            }        }        else             fail[t]=Min[t]=1;        last=t;    }}int b[maxn];int main(){    scanf("%s",sa);    last=1,tail=1;    build(sa);    int hh=strlen(sa);    for(int i=1;i<=tail;i++) cnt[Max[i]]++;    for(int i=1;i<=hh;i++) cnt[i]+=cnt[i-1];    for(int i=1;i<=tail;i++) b[cnt[Max[i]]--]=i;    while(~scanf("%s",sb))    {        int h=strlen(sb);        int p=1;        int len=0;        for(int i=0;i<h;i++)        {            int c=sb[i]-'a';            if(nxt[p][c])                len++,p=nxt[p][c];            else{                while(p&&!nxt[p][c])                    p=fail[p];                if(!p) p=1,len=0;                else                {                    len=Max[p]+1;                    p=nxt[p][c];                }            }            Maxi[p]=max(Maxi[p],len);        }        for(int i=tail;i>=1;i--)        {            int p=b[i];            Mini[p]=min(Mini[p],Maxi[p]);            if(fail[p]) Maxi[fail[p]]=max(Maxi[p],Maxi[fail[p]]);            Maxi[p]=0;        }    }    int ans=0;    for(int i=1;i<=tail;i++)    {        ans=max(ans,Mini[i]);    }    printf("%d",ans );}