SP1811 LCS - Longest Common Substring

Description

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 simple, for two given strings, find the length of the longest common substring of them.

Here common substring means a substring of two or more strings.

Solution

将两个字符串连接成一个，中间加入分隔符，加入后缀自动机

在后缀链接上DP求子串出现次数

如果某个子串在A串和B串都出现过，那么可以用maxlen更新答案

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int tot=1,las=1,len1,len2,cur,siz[1000005][2],head[1000005],tot2,ans;
char s1[250005],s2[250005];
struct SAM
{
    int ch[27],fa,len;
}sam[1000005];
struct Edge
{
    int to,nxt;
}edge[1000005];
inline int read()
{
    int w=0,f=1;
    char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9')w=(w<<1)+(w<<3)+ch-'0',ch=getchar();
    return w*f;
}
void insert(int c)
{
    int p=las,np=las=++tot;
    sam[np].len=sam[p].len+1,siz[np][cur]=1;
    for(;p&&!sam[p].ch[c];p=sam[p].fa) sam[p].ch[c]=np;
    if(!p) sam[np].fa=1;
    else
    {
        int q=sam[p].ch[c];
        if(sam[q].len==sam[p].len+1) sam[np].fa=q;
        else
        {
            int nq=++tot;
            sam[nq]=sam[q],sam[nq].len=sam[p].len+1,sam[np].fa=sam[q].fa=nq;
            for(;p&&sam[p].ch[c]==q;p=sam[p].fa) sam[p].ch[c]=nq;
        }
    }
}
void dfs(int k)
{
    for(int i=head[k];i;i=edge[i].nxt)
    {
        int v=edge[i].to;
        dfs(v),siz[k][0]+=siz[v][0],siz[k][1]+=siz[v][1];
    }
    if(siz[k][0]&&siz[k][1]) ans=max(ans,sam[k].len);
}
int main()
{
    scanf("%s",s1+1),scanf("%s",s2+1),len1=strlen(s1+1),len2=strlen(s2+1);
    for(int i=1;i<=len1;i++) insert(s1[i]-'a');
    insert(26),cur=1;
    for(int i=1;i<=len2;i++) insert(s2[i]-'a');
    for(int i=1;i<=tot;i++) edge[++tot2]=(Edge){i,head[sam[i].fa]},head[sam[i].fa]=tot2;
    dfs(0),printf("%d\n",ans);
    return 0;
}