【后缀自动机】 SPOJ LCS

将A建立成后缀自动机，后缀自动机的任意一个节点都表示若干个A的字串，让B在后缀自动机上匹配，不能匹配就沿着fa指针转移。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 500005
#define maxm 10005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct node
{
	int len;
	node *fa, *next[26];
}*tail, *root, pool[maxn], *last;
char a[maxn];
char b[maxn];

node* newnode(int len)
{
	tail->len = len;
	tail->fa = 0;
	memset(tail->next, 0, sizeof tail->next);
	return tail++;
}

void init()
{
	tail = pool;
	root = newnode(0);
	last = root;
}

void add(int c)
{
	node *p = last, *np = newnode(last->len + 1);
	last = np;
	for(; p && !p->next[c]; p = p->fa) p->next[c] = np;
	if(!p) np->fa = root;
	else {
		node *q = p->next[c];
		if(p->len + 1 == q->len) np->fa = q;
		else {
			node *nq = newnode(q->len);
			*nq = *q;
			nq->len = p->len + 1;
			np->fa = q->fa = nq;
			for(; p && p->next[c] == q; p = p->fa) p->next[c] = nq;
		}
	}
}

void work()
{
	init();
	scanf("%s%s", a, b);
	int len = strlen(a);
	for(int i = 0; i < len; i++) add(a[i] - 'a');
	len = strlen(b);
	int ans = 0, res = 0;
	node *p = root;
	for(int i = 0; i < len; i++) {
		int c = b[i] - 'a';
		if(p->next[c]) res++, p = p->next[c];
		else {
			for(; p && !p->next[c]; p = p->fa);
			if(!p) res = 0, p = root;
			else res = p->len + 1, p = p->next[c];
		}
		ans = max(ans, res);
	}
	printf("%d\n", ans);
}

int main()
{
	work();
	
	return 0;
}