SPOJ 1812 Longest Common Substring II

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

題解：
這題走了許多彎路，最後好不容易拍完錯誤，代碼沒有改得精簡.
思路可以參考：
根據上題可以想到，拿每一個串都在A得SAM上跑，然後記錄當前串在SAM得每一個節點的最大值，然後如果是多個串的話
直接記錄最小值即可，最後掃一邊記錄最小值的最大值即可
注意一些細節：
對於沒一個串的匹配，如果當前結點被匹配到了，那麼他的父親結點都可以匹配到，那麼直接拓撲序更新即可，不然就WA#10哦..

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#define il inline
#define RG register
using namespace std;
const int N=350005;
char s[N];int cur=1,cnt=1,last=1,fa[N],ch[N][27],dis[N],n=0;
void build(int c)
{
    last=cur;cur=++cnt;
    RG int p=last;dis[cur]=++n;
    for(;p && !ch[p][c];p=fa[p])ch[p][c]=cur;
    if(!p)fa[cur]=1;
    else{
        int q=ch[p][c];
        if(dis[q]==dis[p]+1)fa[cur]=q;
        else{
            int nt=++cnt;dis[nt]=dis[p]+1;
            memcpy(ch[nt],ch[q],sizeof(ch[q]));
            fa[nt]=fa[q];fa[q]=fa[cur]=nt;
            for(;ch[p][c]==q;p=fa[p])ch[p][c]=nt;
        }
    }
}
int id=0,t[N],tot[N],mx[N],sa[N];bool mark[N][15];
void FLr()
{
    RG int p=1,l=strlen(s),c,u,len=0;id++;
    for(RG int i=0;i<l;i++){
        c=s[i]-'a';
        u=ch[p][c];
        if(u){
            mark[u][id]=true;len++;
            p=u;
        }
        else{
            while(p && !ch[p][c])p=fa[p];
            if(p){
                mark[ch[p][c]][id]=true;
                len=dis[p]+1;
                p=ch[p][c];
            }
            else p=1,len=0;
        }
        if(len>mx[p])mx[p]=len;
    }
    for(RG int i=cnt;i;i--){
        p=sa[i];
        if(mx[p]<tot[p])tot[p]=mx[p];
        if(fa[p] && mark[p][id])mx[fa[p]]=dis[fa[p]];
        mx[p]=0;
    }
}
int c[N];
void Dfp(){
    RG int p,i,j;
    for(i=cnt;i;i--){
        p=sa[i];
        for(j=1;j<=id;j++)
            mark[fa[p]][j]|=mark[p][j];
    }
    for(i=1;i<=cnt;i++)
        for(j=1;j<=id;j++)
            if(mark[i][j])t[i]++;
}
int ans=0;
void dfs(){
    for(int i=2;i<=cnt;i++){
        if(t[i] && tot[i]>ans)ans=tot[i];
    }
}
void jip(){
    RG int i;
    for(i=1;i<=cnt;i++)c[dis[i]]++;
    for(i=1;i<=n;i++)c[i]+=c[i-1];
    for(i=cnt;i;i--)sa[c[dis[i]]--]=i;
}
void work()
{
    scanf("%s",s);
    for(RG int i=0,l=strlen(s);i<l;i++)build(s[i]-'a');
    for(RG int i=1;i<=cnt;i++)tot[i]=N;
    jip();
    while(~scanf("%s",s))FLr();
    Dfp();dfs();
    printf("%d
",ans);
}
int main()
{
    freopen("pow.in","r",stdin);
    freopen("pow.out","w",stdout);
      work();
    return 0;
}