SPOJ LCS2 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

Notice: new testcases added

就是求多个字符串的最长公共子串

后缀自动机

用第一个串建后缀自动机，然后在这个自动机上花式转移。

solve()里面tmp没初始化，调了半天……

/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int mxn=200010;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct SAM{
    int t[mxn][27],l[mxn];
    int fa[mxn];
    int S,last,cnt;
    int rk[mxn],w[mxn],mx[mxn],mn[mxn];
    void init(){S=last=cnt=1;return;}
    void add(int c){
        int p=last,np=++cnt;last=np;
        l[np]=l[p]+1;
        for(;p && !t[p][c];p=fa[p])t[p][c]=np;
        if(!p)
            fa[np]=S;
        else{
            int q=t[p][c];
            if(l[p]+1==l[q])
                fa[np]=q;
            else{
                int nq=++cnt;
                l[nq]=l[p]+1;
                memcpy(t[nq],t[q],sizeof t[q]);
                fa[nq]=fa[q];
                fa[q]=fa[np]=nq;
                for(;t[p][c]==q;p=fa[p])t[p][c]=nq;
            }
        }
    }
    void st(int len){
        for(int i=1;i<=cnt;i++){
            mn[i]=l[i];    w[l[i]]++;
        }
        for(int i=1;i<=len;i++)w[i]+=w[i-1];
        for(int i=1;i<=cnt;i++)rk[w[l[i]]--]=i;
        return;
    }
    void solve(char *s){
        int now=S,len=strlen(s+1),tmp=0;
        memset(mx,0,sizeof mx);
        int i,j;
        for(i=1;i<=len;i++){
            int c=s[i]-'a';
            if(t[now][c]){tmp++;now=t[now][c];}
            else{
                while(now && !t[now][c])now=fa[now];
                if(!now){tmp=0;now=S;}
                else{tmp=l[now]+1;now=t[now][c];}
            }
            mx[now]=max(mx[now],tmp);
        }
        for(i=cnt;i;i--){
            now=rk[i];
            if(mx[now]<mn[now])mn[now]=mx[now];//多串对应位置取最小值
            if(fa[now] && mx[now])mx[fa[now]]=max(mx[fa[now]],mx[now]);
        }
    }
}sa;
char s[mxn];
int main(){
    int i,j;
    scanf("%s",s+1);
    int len=strlen(s+1);
    sa.init();
    for(i=1;i<=len;i++)sa.add(s[i]-'a');
    sa.st(len);
    while(scanf("%s",s+1)!=EOF){
        sa.solve(s);
    }
    int ans=0;
    for(i=1;i<=sa.cnt;i++){
        ans=max(ans,sa.mn[i]);
    }
    printf("%d
",ans);
    return 0;
}

（没放代码就保存了页面，过了好久才发现，也是蠢）