UVA 12345 Dynamic len(带修莫队)

Dynamic len

【题目链接】Dynamic len

【题目类型】带修莫队

&题解：

莫队可以单点更改,只要再多加一维,代表查询次数,排序的时候3个关键字.
之后循环离线的时候,先暴力时间指针(oi大佬说的),之后l,和r就随便了.还有要会用vis数组.

【时间复杂度】\(O(n^{\frac{5}{3}})\)

&代码：

#include <cstdio>
#include <bitset>
#include <iostream>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define ll long long
const int maxn = 5e4 + 7;
struct nt {
	int l, r, id, ts;
} ask[maxn];
struct nd {
	int x, pre, v;
} p[maxn];
int n, m, an[maxn], c[maxn], pos[maxn], num[maxn * 20], last[maxn], ans;
bool vis[maxn];
char s[9];
bool cmp(nt a, nt b) {
	return pos[a.l] < pos[b.l] ||
	       pos[a.l] == pos[b.l] && pos[a.r] < pos[b.r] ||
	       pos[a.l] == pos[b.l] && pos[a.r] == pos[b.r] && a.ts < b.ts;
}
void update(int x) {
	if(vis[x]) {
		if(!--num[c[x]]) --ans;
	}
	else if(++num[c[x]] == 1) ans++;
	vis[x] ^= 1;
}
void change(int x, int v) {
	if(vis[x]) {
		update(x); c[x] = v; update(x);
	}
	else c[x] = v;
}
int main() {
	//("E:1.in", "r", stdin);
	scanf("%d%d", &n, &m);
	int x, y, tot = 0, top = 0;
	int bk = sqrt(n);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &c[i]);
		last[i] = c[i];
		pos[i] = i / bk;
	}
	for(int i = 1; i <= m; i++) {
		scanf("%s%d%d", s, &x, &y);
		x++;
		if(s[0] == 'Q') {
			//[++tot].id必须要赋为tot. 而且这句话必须在第一位写
			ask[++tot].id = tot;
			ask[tot].l = x, ask[tot].r = y, ask[tot].ts = top;
		}
		else {
			p[++top].x = x, p[top].v = y, p[top].pre = last[x];
			last[x] = y;
		}
	}
	sort(ask + 1, ask + 1 + tot, cmp);
	int now = 0, pl = 1, pr = 0;
	for(int i = 1; i <= tot; i++) {
		if(now < ask[i].ts) {
			for(int j = now + 1; j <= ask[i].ts; j++)
				change(p[j].x, p[j].v);
		}
		else {
			for(int j = now; j >= ask[i].ts + 1; j--)
				change(p[j].x, p[j].pre);
		}
		if(pr < ask[i].r) {
			for(int j = pr + 1; j <= ask[i].r; j++)
				update(j);
		}
		else {
			for(int j = pr; j >= ask[i].r + 1; j--)
				update(j);
		}
		if(pl < ask[i].l) {
			for(int j = pl; j <= ask[i].l - 1; j++)
				update(j);
		}
		else {
			for(int j = pl - 1; j >= ask[i].l; j--)
				update(j);
		}
		pl = ask[i].l, pr = ask[i].r, now = ask[i].ts;
		an[ask[i].id] = ans;
	}
	for(int i = 1; i <= tot; i++) {
		printf("%d\n", an[i]);
	}
	return 0;
}