做法:
- 后缀家族的基本应用,之前学习了后缀数组的解法,后缀自动机的思想如下:
- 给第一个串建立自动机,第二个串在自动机上匹配,通过tran数组,如果匹配就len+1,
否则通过slink数组(fa)向前跳,直到能继续匹配。
AC代码:
#include<bits/stdc++.h>
using namespace std;
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x) push_back(x)
#define sz(x) (int)(x).size()
#define sc(x) scanf("%d",&x)
#define abs(x) ((x)<0 ? -(x) : x)
#define all(x) x.begin(),x.end()
#define mk(x,y) make_pair(x,y)
#define fin freopen("in.txt","r",stdin)
#define fout freopen("out.txt","w",stdout)
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 3e5+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;
char s[maxn],b[maxn];
struct SuffixAutoMation{
int root,cnt,last,len;
int trans[maxn<<1][26],slink[maxn<<1],l[maxn<<1];
inline void init()
{
root = cnt = last = 1;
memset(trans,0,sizeof(trans));
slink[1] = l[1] = 0;
}
inline void add(int x)
{
int p = last,np = ++cnt;last = np;l[np] = l[p]+1;
for(;p && !trans[p][x];p = slink[p]) trans[p][x] = np;
if(!p) slink[np] = 1;
else{
int q = trans[p][x];
if(l[p]+1 == l[q]) slink[np] = q;
else{
int nq = ++cnt;l[nq] = l[p]+1;
memcpy(trans[nq],trans[q],sizeof(trans[q]));
slink[nq] = slink[q];
slink[q] = slink[np] = nq;
for(;trans[p][x] == q; q = slink[q]) trans[p][x] = nq;
}
}
}
inline void build()
{
init();
scanf("%s",s+1);
scanf("%s",b+1);
len = strlen(s+1);
for(int i=1;i<=len;i++) add(s[i]-'a');
}
inline int Find() //第二个串与第一个串匹配
{
int m = strlen(b+1);
int res = 0,ans = 0,p = root;
for(int i=1;i<=m;i++)
{
int x = b[i]-'a';
if(trans[p][x]) res++,p = trans[p][x];
else{
for(;p && !trans[p][x];p = slink[p]);
if(!p) res = 0,p = root;
else res = l[p]+1, p = trans[p][x];
}
ans = max(ans,res);
}
return ans;
}
}sam;
int main()
{
// fin;
// IO;
sam.build();
cout<<sam.Find()<<endl;
return 0;
}