Description
维护一个字符串,支持插入字符,修改字符,以及求两个后缀的\(lcp\)。
Solution
建立一棵\(Splay\)来维护整个串,每个节点维护整个子树的哈希值。对于插入,直接在对应的位置插入;修改也直接修改就好;然后一路\(update\)。对于查询,考虑二分,然后每次查询对应区间的哈希值,\(O(1)\)判断即可。
查询某个区间的哈希值,用\(Splay\)写的话比较方便,对于区间\([l,r]\),直接把\(l-1\)旋转到根,把\(r+1\)旋转到根节点的右儿子,那么根节点的右儿子的左儿子的哈希值就是区间\([l,r]\)的哈希值。
Code
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
const int _ = 1e5 + 10;
const int __ = 3e5 + 10;
const int base = 233;
char s[_];
int N, M;
ull p[__];
struct Splay {
int fa[__], son[__][2], val[_], siz[_];
ull h[_];
int tot, root;
int alloc(int v, int f) {
fa[++tot] = f, siz[tot] = 1;
son[tot][0] = son[tot][1] = 0;
val[tot] = h[tot] = v;
return tot;
}
void update(int x) {
siz[x] = siz[son[x][0]] + siz[son[x][1]] + 1;
h[x] = h[son[x][1]] + val[x] * p[siz[son[x][1]]] +
h[son[x][0]] * p[siz[son[x][1]] + 1];
}
int ident(int x) { return son[fa[x]][1] == x; }
void connect(int x, int f, int k) { fa[x] = f, son[f][k] = x; }
void rotate(int x) {
int y = fa[x], z = fa[y];
if (y == root) root = x;
int yson = ident(x), zson = ident(y);
int k = son[x][yson ^ 1];
connect(k, y, yson);
connect(y, x, yson ^ 1);
connect(x, z, zson);
update(y), update(x);
}
void splay(int x, int to) {
while (fa[x] != to) {
int y = fa[x], z = fa[y];
if (z != to) (ident(x)) ^ (ident(y)) ? rotate(x) : rotate(y);
rotate(x);
}
if (!to) root = x;
}
int find(int x) {
int u = root;
while (1) {
if (x == siz[son[u][0]] + 1) return u;
if (x <= siz[son[u][0]])
u = son[u][0];
else
x -= siz[son[u][0]] + 1, u = son[u][1];
}
}
void insert(int x, int v) {
if (!root) {
root = alloc(v, 0);
return;
}
int u = find(x);
splay(u, 0);
if (!son[u][1]) {
son[u][1] = alloc(v, u);
update(u);
return;
} else {
u = find(x + 1);
splay(u, root);
son[u][0] = alloc(v, u);
update(u), update(root);
return;
}
}
void modify(int x, int v) {
int u = find(x);
splay(u, 0);
val[u] = v;
update(u);
}
ull query(int l, int r) {
int u = find(l);
splay(u, 0);
if (l == r) return val[u];
u = find(r);
splay(u, root);
return h[son[u][0]] * p[1] + val[u] + val[root] * p[siz[son[u][0]] + 1];
}
} tr;
inline void init(const int lim = 3e5) {
p[0] = 1;
for (int i = 1; i <= lim; ++i) p[i] = p[i - 1] * base;
for (int i = 1; i <= N; ++i) tr.insert(i - 1, s[i] - 'a' + 1);
}
inline bool check(int x, int y, int len) {
return tr.query(x, x + len - 1) == tr.query(y, y + len - 1);
}
int calc(int x, int y) {
int l = 0, r = min(N - x, N - y) + 1;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (check(x, y, mid))
l = mid;
else
r = mid - 1;
}
return l;
}
int main() {
#ifndef ONLINE_JUDGE
freopen("martian.in", "r", stdin);
freopen("martian.out", "w", stdout);
#endif
scanf("%s%d", s + 1, &M);
N = strlen(s + 1);
init();
char op[3], v[3];
int x, y;
while (M--) {
scanf("%s%d", op, &x);
if (op[0] == 'Q') {
scanf("%d", &y);
printf("%d\n", calc(x, y));
} else if (op[0] == 'R') {
scanf("%s", v);
char c = v[0];
tr.modify(x, c - 'a' + 1);
} else if (op[0] == 'I') {
scanf("%s", v);
char c = v[0];
tr.insert(x, c - 'a' + 1);
++N;
}
}
return 0;
}