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=;
 char s[N];int cur=,cnt=,last=,fa[N],ch[N][],dis[N],n=;
 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]=;
     else{
         int q=ch[p][c];
         if(dis[q]==dis[p]+)fa[cur]=q;
         else{
             int nt=++cnt;dis[nt]=dis[p]+;
             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=,t[N],tot[N],mx[N],sa[N];bool mark[N][];
 void FLr()
 {
     RG int p=,l=strlen(s),c,u,len=;id++;
     for(RG int i=;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]+;
                 p=ch[p][c];
             }
             else p=,len=;
         }
         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]=;
     }
 }
 int c[N];
 void Dfp(){
     RG int p,i,j;
     for(i=cnt;i;i--){
         p=sa[i];
         for(j=;j<=id;j++)
             mark[fa[p]][j]|=mark[p][j];
     }
     for(i=;i<=cnt;i++)
         for(j=;j<=id;j++)
             if(mark[i][j])t[i]++;
 }
 int ans=;
 void dfs(){
     for(int i=;i<=cnt;i++){
         if(t[i] && tot[i]>ans)ans=tot[i];
     }
 }
 void jip(){
     RG int i;
     for(i=;i<=cnt;i++)c[dis[i]]++;
     for(i=;i<=n;i++)c[i]+=c[i-];
     for(i=cnt;i;i--)sa[c[dis[i]]--]=i;
 }
 void work()
 {
     scanf("%s",s);
     for(RG int i=,l=strlen(s);i<l;i++)build(s[i]-'a');
     for(RG int i=;i<=cnt;i++)tot[i]=N;
     jip();
     while(~scanf("%s",s))FLr();
     Dfp();dfs();
     printf("%d\n",ans);
 }
 int main()
 {
     freopen("pow.in","r",stdin);
     freopen("pow.out","w",stdout);
       work();
     return ;
 }