spoj LCS 【后缀自动机】

琦不会后缀自动机……

是以前太浪了……

所以所有东西都留到了noi前来学……

马上狗牌退役了TAT（心塞qwq

题目大意：给出两个串A,B，求A、B的最长公共子串

对A建后缀自动机，然后用B去匹配，若能匹配上就转移到儿子，否则沿着parent树向上跳

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define N 500005
using namespace std;

int n,m,last = 1,tot = 1,p,q,np,nq,ans;
int son
[26],par
,mx
;
char a
,b
;

int new_node(int x)
{
mx[++ tot] = x;
return tot;
}

void add(int x)
{
p = last;
np = new_node(mx[p] + 1);
for (;p && !son[p][x];p = par[p]) son[p][x] = np;
if (!p) par[np] = 1;
else
{
q = son[p][x];
if (mx[q] == mx[p] + 1) par[np] = q;
else
{
nq = new_node(mx[p] + 1);
memcpy(son[nq],son[q],sizeof(son[nq]));
par[nq] = par[q],par[q] = par[np] = nq;
for (;son[p][x] == q;p = par[p]) son[p][x] = nq;
}
}
last = np;
}

int main()
{
scanf("%s%s",a + 1,b + 1);
n = strlen(a + 1),m = strlen(b + 1);
for (int i = 1;i <= n;i ++) add(a[i] - 'a');
for (int i = 1,p = 1,l = 0;i <= m;i ++)
{
int x = b[i] - 'a';
if (son[p][x]) l ++,p = son[p][x];
else
{
for (;p && !son[p][x];p = par[p]);
if (!p) l = 0,p = 1;
else l = mx[p] + 1,p = son[p][x];
}
ans = max(ans,l);
}
cout << ans << endl;

return 0;
}