【后缀自动机】 CodeForces 235C Cyclical Quest

先对原串建立后缀自动机。考虑每个询问。询问的子串相当于考虑它的n个循环串，因此我们把询问串连接在询问串，那么这个新串的长度为n的子串就是所要求的串。把新串放在后缀自动机上面跑，同时记录现在新串匹配的最长后缀len。如果len>=当前串长度n，那么我们就沿着fa指针跳到满足len>=n且长度最小的节点。那么原点到这个节点必然有一条路径的字符串是当前匹配的长度为n的后缀。这个节点的right集合大小就是所求的答案。但是可能会出现重复，每个节点记录一个vis值就可以了。。累加答案即可。。。

#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 1000005
#define maxm 2000005
#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

const int alpha = 26;

struct node
{
	int len, cnt, vis;
	node *ch[alpha], *fa;
}*last, *tail, pool[maxm], *root;

node *tp[maxm];
char ss[maxm];
char s[maxm];
int c[maxm];
int n, m;

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

void init()
{
	tail = pool;
	root = last = newnode(0);
	memset(c, 0, sizeof c);
}

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

void solve(int T)
{
	int ans = 0;
	node *p = root;
	int len = 0;
	int n = strlen(s);
	for(int i = n; i < 2 * n; i++) s[i] = s[i-n];
	for(int i = 0; i < 2 * n; i++) {
		int t = s[i] - 'a';
		while(p && !p->ch[t]) {
			p = p->fa;
			if(p) len = p->len;
			else len = 0;
		}
		if(p) p = p->ch[t], len++;
		else p = root, len = 0;
		if(len >= n) {
			while(p->fa->len >= n) p = p->fa, len = p->len;
			if(p->vis != T) p->vis = T, ans += p->cnt;
		}
	}
	printf("%d\n", ans);
}

void work()
{
	scanf("%s", ss);
	init();
	for(int i = 0; ss[i]; i++) add(ss[i] - 'a');
	node *o = root;
	for(int i = 0; ss[i]; i++) o = o->ch[ss[i] - 'a'], o->cnt++;
	int n = strlen(ss);
	for(node *p = pool; p != tail; p++) c[p->len]++;
	for(int i = 1; i <= n; i++) c[i] += c[i-1];
	for(node *p = pool; p != tail; p++) tp[--c[p->len]] = p;
	int tot = tail - pool;
	for(int i = tot - 1; i > 0; i--) tp[i]->fa->cnt += tp[i]->cnt;

	scanf("%d", &m);
	for(int i = 1; i <= m; i++) {
		scanf("%s", s);
		solve(i);
	}
}

int main()
{
	work();

	return 0;
}