【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;}// headconst 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;}