[洛谷P1903][国家集训队]数颜色

题目大意：有$n$支画笔，有两个操作

1. $Q;l;r：$询问$[l,r]$中有几种颜色
2. $R;p;Col：$把第$p$支画笔的颜色改成$Col$

题解：带修莫队，分为$n^{frac{1}{3}}$个块，每个块$n^{frac{2}{3}}$个元素，复杂度$O(n^{frac{5}{3}})$

卡点：读入时，修改的时间赋值错

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 50010
#define bl(x) (x >> 10)
int n, m, Qcnt, Mcnt;
int C[maxn], ans[maxn], cnt[1000010];
struct Query {
	int l, r, t, id;
	inline bool operator < (const Query &rhs) const {
		return bl(l) == bl(rhs.l) ? (bl(r) == bl(rhs.r) ? t < rhs.t : r < rhs.r) : l < rhs.l;
	}
} Q[maxn];
struct Modify {
	int pos, t, col;
} M[maxn];
int l = 0, r = 0, p = 0, res = 0;
inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
void modify(int x) {
	if (M[x].pos >= l && M[x].pos <= r) {
		res -= --cnt[C[M[x].pos]] == 0;
		res += cnt[M[x].col]++ == 0;
	}
	swap(M[x].col, C[M[x].pos]);
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &C[i]);
	for (int i = 1; i <= m; i++) {
		char op[10];
		scanf("%s", op);
		if (*op == 'Q') {
			Qcnt++;
			scanf("%d%d", &Q[Qcnt].l, &Q[Qcnt].r);
			Q[Qcnt].t = i;
			Q[Qcnt].id = Qcnt;
		} else {
			Mcnt++;
			scanf("%d%d", &M[Mcnt].pos, &M[Mcnt].col);
			M[Mcnt].t = i;
		}
	}
	std::sort(Q + 1, Q + Qcnt + 1);
	for (int i = 1; i <= Qcnt; i++) {
		while (l > Q[i].l) res += cnt[C[--l]]++ == 0;
		while (r < Q[i].r) res += cnt[C[++r]]++ == 0;
		while (l < Q[i].l) res -= --cnt[C[l++]] == 0;
		while (r > Q[i].r) res -= --cnt[C[r--]] == 0;
		while (p < Mcnt && M[p + 1].t <= Q[i].t) modify(++p);
		while (p && M[p].t > Q[i].t) modify(p--);
		ans[Q[i].id] = res;
	}
	for (int i = 1; i <= Qcnt; i++) printf("%d
", ans[i]);
	return 0;
}