洛谷 P4036 [JSOI2008]火星人（splay+字符串hash）

题面

洛谷

题解

首先，我们知道求最长公共前缀可以用二分答案+hash来求

因为有修改操作， 考虑将整个字符串的hash值放入splay中

接着就是splay的基本操作了

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register
const int N = 500010;
typedef unsigned long long ull;
using namespace std;
inline int gi() {
	RG int x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();
	return f ? -x : x;
}

ull p[N];
int tot;
struct Splay {
	int fa, ch[2], siz, v;
	ull hs;
}t[N];
int rt;
#define get(x) (t[t[x].fa].ch[1] == x)
#define ls t[x].ch[0]
#define rs t[x].ch[1]
inline void pushup(int x) {
	t[x].siz = t[ls].siz + t[rs].siz + 1;
	t[x].hs = t[rs].hs + (ull)t[x].v*p[t[rs].siz] + t[ls].hs*p[t[rs].siz+1];
}
void build(int l, int r, int x) {
	if (l > r) return ;
	int mid = (l + r) >> 1;
	if (mid >= x) rs = mid;
	else ls = mid;
	t[mid].fa = x, t[mid].siz = 1;
	if (l == r) return ;
	build(l, mid-1, mid), build(mid+1, r, mid);
	pushup(mid);
	return ;
}
void rotate(int x) {
	int y = t[x].fa, z = t[y].fa, k = get(x);
	t[x].fa = z; t[z].ch[get(y)] = x;
	t[t[x].ch[k^1]].fa = y; t[y].ch[k] = t[x].ch[k^1];
	t[y].fa = x; t[x].ch[k^1] = y;
	pushup(y);
	return ;
}
void splay(int x, int goal) {
	while (t[x].fa != goal) {
		int y = t[x].fa, z = t[y].fa;
		if (z != goal)
			(get(x)^get(y)) ? rotate(x) : rotate(y);
		rotate(x);
	}
	pushup(x);
	if (!goal) rt = x;
}
int kth(int k) {
	int x = rt;
	while (233) {
		if (t[ls].siz+1 == k) return x;
		if (t[ls].siz+1 < k) k -= (t[ls].siz+1), x = rs;
		else x = ls;
	}
}
ull get_hash(int l, int r) {
	int x = kth(l), y = kth(r+2);
	splay(x, 0); splay(y, rt);
	return t[t[t[rt].ch[1]].ch[0]].hs;
}
char ch[N];
int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	p[0] = 1;
	for (int i = 1; i <= 300000; i++) p[i] = (ull)27*p[i-1];
	scanf("%s", ch+1);
	int n = strlen(ch+1);
	for (int i = 2; i <= n+1; i++)
		t[i].v = t[i].hs = ch[i-1]-'a'+1;
	build(1, n+2, rt);
	rt = (n+3) >> 1;
	tot = n+2;
	int T = gi();
	while (T--) {
		char s[5];
		scanf("%s", s);
		if (s[0] == 'Q') {
			int x = gi(), y = gi();
			if (x > y) swap(x, y);
			int l = 1, r = tot-y-1;
			while (l <= r) {
				int mid = (l + r) >> 1;
				if (get_hash(x, x+mid-1) == get_hash(y, y+mid-1))
					l = mid+1;
				else r = mid-1;
			}
			printf("%d\n", l-1);
		}
		else if (s[0] == 'R') {
			int x = gi();
			scanf("%s", s);
			splay(kth(x+1), 0);
			t[rt].hs -= (ull)t[rt].v*p[t[t[rt].ch[1]].siz];
			t[rt].v = s[0]-'a'+1;
			t[rt].hs += (ull)t[rt].v*p[t[t[rt].ch[1]].siz];			
		}
		else {
			int x = gi();
			scanf("%s", s);
			int k1 = kth(x+1), k2 = kth(x+2);
			splay(k1, 0); splay(k2, k1);
			t[t[rt].ch[1]].ch[0] = ++tot;
			t[tot].fa = t[rt].ch[1];
			t[tot].v = t[tot].hs = s[0]-'a'+1;
			splay(tot, 0);
		}
	}
	return 0;
}