luogu P4688 [Ynoi2016] 掉进兔子洞

https://www.luogu.com.cn/problem/P4688

思路还是能一眼看出来的，主要是有一些小trick
离散化的时候不要去重（填数的时候往后补就好了）
然后发现空间回炸
把询问分个组就好了
10000一组来做，时间复杂度不变，空间缩小
岂不美哉
具体看代码吧
code:

#include<bits/stdc++.h>
#define N 100005
using namespace std;
struct Q {
	int l, r, id;
} q[N << 2];
int bel[N];
int cmp(Q x, Q y) {
	if(bel[x.l] != bel[y.l]) return x.l < y.l;
	return x.r < y.r;
}
bitset<N> mi[20005], s;
int gs[N], vis[N], a[N], b[N], ans[N];
void add(int x) { x = a[x];
	s[x + gs[x]] = 1;//往后填
	gs[x] ++;
}
void del(int x) { x = a[x];
	gs[x] --;
	s[x + gs[x]] = 0;	
}
void solve(int m) {
	memset(gs, 0, sizeof gs);
	for(int i = 0; i <= m; i ++) vis[i] = ans[i] = 0;
	int sz = 0;
	for(int i = 1; i <= m; i ++) 
		for(int j = 1; j <= 3; j ++)
			sz ++, scanf("%d%d", &q[sz].l, &q[sz].r), q[sz].id = i, ans[i] += q[sz].r - q[sz].l + 1;
	sort(q + 1, q + 1 + sz, cmp);
	int l = 0, r = -1;
	s.reset();
	for(int i = 1; i <= sz; i ++) {	
		while(r < q[i].r) add(++ r);
		while(l > q[i].l) add(-- l);
		while(r > q[i].r) del(r --);
		while(l < q[i].l) del(l ++);
		if(vis[q[i].id]) mi[q[i].id] &= s;
		else mi[q[i].id] = s;
		vis[q[i].id] = 1;
	}
	for(int i = 1; i <= m; i ++) printf("%d\n", ans[i] - 3 * mi[i].count());
}
int n, m;
int main() {	
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i ++) scanf("%d", &a[i]), b[i] = a[i];
	sort(b + 1, b + 1 + n);
	for(int i = 1; i <= n; i ++) a[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
	int blo = sqrt(n);
	for(int i = 1; i <= n; i ++) bel[i] = (i - 1) / blo + 1;
	int mm = 10000;
	while(m) {
		solve(min(mm, m));//分组做
		m -= min(mm, m);
	}
	return 0;
}

空间开不下可以分组做，记住这个小trick