看了《Hash在信息学竞赛中的一类应用》中的例题3,这道题很类似啊,只不过没有删点和区间翻转。
用Splay维护字符串哈希,加点改点什么的就不用说了,查询时二分答案,这样时间复杂度是$O(mlog^2 n)$的
论文的例题3中删点很简单,和插点一样,不用说了,区间翻转只要打一个翻转标记,维护正序hash和逆序hash,翻转时交换两个hash值即可。
对拍终于成功了QAQ,插点时孩子不认父亲TwT又手残了~
《Hash在信息学竞赛中的一类应用》中还提到了块状链表的做法,都很易懂:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100003;
const int p = 9875321; int P[N];
struct node *null;
struct node {
node *ch[2], *fa;
int k, s, ha;
node (int _k = 0) {k = _k; s = 1; ha = _k; ch[0] = ch[1] = fa = null;}
void setc(node *r, bool c) {this->ch[c] = r; r->fa = this;}
bool pl() {return fa->ch[1] == this;}
void count() {
s = ch[0]->s + ch[1]->s + 1;
ha = ((ch[0]->ha + 1ll * k * P[ch[0]->s]) % p + 1ll * ch[1]->ha * P[ch[0]->s + 1] % p) % p;
}
} *root; int n, m;
char s[N]; namespace Splay {
node *Build(int l, int r) {
if (l > r) return null;
int mid = (l + r) >> 1;
node *t = new node(s[mid] - 'a');
t->ch[0] = Build(l, mid - 1); t->setc(t->ch[0], 0);
t->ch[1] = Build(mid + 1, r); t->setc(t->ch[1], 1);
t->count();
return t;
}
void init() {
P[0] = 1; for(int i = 1; i < N; ++i) P[i] = P[i - 1] * 26 % p;
null = new node; null->s = 0; null->ch[0] = null->ch[1] = null->fa = null;
scanf("%s", s + 1); n = strlen(s + 1); root = Build(1, n);
}
void rotate(node *r) {
node *f = r->fa;
bool c = r->pl();
if (f != root) f->fa->setc(r, f->pl());
else root = r, r->fa = null;
f->setc(r->ch[!c], c);
r->setc(f, !c);
f->count();
}
void splay(node *r, node *tar = null) {
for(; r->fa != tar; rotate(r))
if (r->fa->fa != tar) rotate(r->pl() == r->fa->pl() ? r->fa : r);
r->count();
}
node *kth(int k) {
node *r = root;
while (1) {
if (r->ch[0]->s >= k) r = r->ch[0];
else if (r->ch[0]->s + 1 >= k) return r;
else k -= (r->ch[0]->s + 1), r = r->ch[1];
}
}
int hash(int l, int r) {
if (l == 1 && r == n) return root->ha;
else if (l == 1) {splay(kth(r + 1)); return root->ch[0]->ha;}
else if (r == n) {splay(kth(l - 1)); return root->ch[1]->ha;}
else {splay(kth(l - 1)); splay(kth(r + 1), root); return root->ch[1]->ch[0]->ha;}
}
void QQ(int l, int r) {
int left = 0, right = root->s - max(l, r) + 1, mid;
while (left < right) {
mid = (left + right + 1) >> 1;
if (hash(l, l + mid - 1) == hash(r, r + mid - 1)) left = mid;
else right = mid - 1;
}
printf("%d\n", left); return;
}
void RR(int k, int num) {
node *r = kth(k);
r->k = num;
splay(r);
}
void II(int k, int num) {
if (k == 0) {
node *r = root;
while (r->ch[0] != null) r = r->ch[0];
r->setc(new node(num), 0);
splay(r->ch[0]);
} else {
splay(kth(k));
if (k == n) {
root->setc(new node(num), 1);
splay(root->ch[1]);
} else {
splay(kth(k + 1), root);
root->ch[1]->setc(new node(num), 0);
splay(root->ch[1]->ch[0]);
}
}
++n;
}
} int main() {
Splay::init();
scanf("%d", &m);
char c; int x, y;
while (m--) {
for(c = getchar(); c < 'A' || c > 'Z'; c = getchar());
switch (c) {
case 'Q':
scanf("%d%d", &x, &y);
Splay::QQ(x, y);
break;
case 'R':
scanf("%d", &x); for(c = getchar(); c < 'a' || c > 'z'; c = getchar());
Splay::RR(x, c - 'a');
break;
case 'I':
scanf("%d", &x); for(c = getchar(); c < 'a' || c > 'z'; c = getchar());
Splay::II(x, c - 'a');
break;
}
}
return 0;
}
对拍大法好~