求两个字符串最长公共子串的长度。
对于第一个串S,建立SAM,之后对于第二个串T,我们在上面和S进行匹配。首先从(t_0)开始,如果能成功匹配的话,那么我们让长度+1,同时更新答案。如果失配,那我们就跳parent树转移到其父节点的位置,并且把当前匹配长度变为其最长后缀长度即可。 最后统计一下输出匹配的最大长度。
#include<bits/stdc++.h>
#define rep(i,a,n) for(register int i = a;i <= n;i++)
#define per(i,n,a) for(register int i = n;i >= a;i--)
#define enter putchar('
')
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define sc second
using namespace std;
typedef long long ll;
const int M = 500005;
const int N = 10000005;
int read()
{
int ans = 0,op = 1;char ch = getchar();
while(ch < '0' || ch > '9') {if(ch == '-') op = -1;ch = getchar();}
while(ch >='0' && ch <= '9') ans = ans * 10 + ch - '0',ch = getchar();
return ans * op;
}
char s[M],t[M];
int n,m,ans,pos;
struct SuffixAutomaton
{
int last,cnt,ch[M<<1][26],fa[M<<1],l[M<<1];
void extend(int c)
{
int p = last,np = ++cnt;
last = cnt,l[np] = l[p] + 1;
while(p && !ch[p][c]) ch[p][c] = np,p = fa[p];
if(!p) {fa[np] = 1;return;}
int q = ch[p][c];
if(l[q] == l[p] + 1) fa[np] = q;
else
{
int nq = ++cnt;
l[nq] = l[p] + 1,memcpy(ch[nq],ch[q],sizeof(ch[q]));
fa[nq] = fa[q],fa[np] = fa[q] = nq;
while(ch[p][c] == q) ch[p][c] = nq,p = fa[p];
}
}
void match(char *a)
{
int v = 1,len = 0;
//rep(i,0,m-1) printf("%c ",a[i]);enter;
rep(i,0,m-1)
{
//printf("#%d %d %d
",v,len,pos);
int c = a[i] - 'a';
while(!ch[v][c] && v != 1) v = fa[v],len = l[v];
if(ch[v][c])
{
v = ch[v][c],len++;
if(len > ans) ans = len;
}
}
}
}SAM;
int main()
{
scanf("%s",s+1),n = strlen(s+1);
scanf("%s",t+1),m = strlen(t+1);
SAM.cnt = SAM.last = 1;
rep(i,1,n) SAM.extend(s[i] - 'a');
SAM.match(t+1);
printf("%d
",ans);
return 0;
}