思路有点奇怪，考虑把查询区间拆分成若干极大区间，最后用O(log^2 n)暴力统计答案，没有TLE，请问是思路太离谱了吗

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define mid ((l + r) >> 1)
#define lch (R << 1)
#define rch ((R << 1) | 1)
ll All[2000005], Max[2000005], F[2000005], E[2000005], N, T, cnt, val[500005], cho[500005];
void Pushup(ll R) {
	All[R] = All[lch] + All[rch];
	Max[R] = max(max(Max[lch], Max[rch]), E[lch] + F[rch]);
	F[R] = max(F[lch], All[lch] + F[rch]);
	E[R] = max(E[rch], All[rch] + E[rch]);
}
void Build(ll R, ll l, ll r) {
	if(l == r){Max[R] = F[R] = E[R] = All[R] = val[l]; return;}
	Build(lch, l, mid);
	Build(rch, mid + 1, r);
	Pushup(R);
}
void Update(ll R, ll l, ll r, ll k, ll v) {
	if(l == r) {
		Max[R] = F[R] = E[R] = All[R] = v;
		return;
	}
	if(k <= mid) Update(lch, l, mid, k, v);
	else Update(rch, mid + 1, r, k, v);
	Pushup(R);
}
void Query(ll R, ll l, ll r, ll ql, ll qr) {
	if(l >= ql && r <= qr) {
		cho[++cnt] = R;
		return;	
	}
	if(mid >= ql) Query(lch, l, mid, ql, qr);
	if(mid + 1 <= qr) Query(rch, mid + 1, r, ql, qr);
}
ll Ans() {
	ll ans = -1e15;
	for(int i = 1; i <= cnt; i++) ans = max(ans, Max[cho[i]]);
	for(int i = 1; i < cnt; i++) {
		ll Sum = 0;
		for(int j = i + 1; j <= cnt; j++) {
			ans = max(ans, E[cho[i]] + Sum + F[cho[j]]);
			Sum += All[cho[j]];
		}
	}
	return ans;
}
int main()
{
	cin >> N >> T;
	for(int i = 1; i <= N; i++) cin >> val[i];
	Build(1, 1, N);
	while(T--) {
		cnt = 0;
		ll op, x, y;
		cin >> op >> x >> y;
		if(op & 1) {
			if(x > y) swap(x, y);
			Query(1, 1, N, x, y);
			cout << Ans() <<'\n';
		}
		else {
			Update(1, 1, N, x, y);
			
		}
	}
	
	
	
	
	return 0;
}