问题一 别忘了建树运行build，不写会出现val = 21934920813的问题 问题二 合并查找函数把三种情况写全 问题三 输进来俩数注意顺序

#include <bits/stdc++.h>
using namespace std;
int n,m;
struct P{
	int maxl,max,maxr,sum;
};
P a[2000005];
int c[500005];
void cut(P t)
{
    cout << t.max << " " << t.maxl << " " << t.maxr << endl;
} 
P pushup(P A,P B)
{
	P x;
	x.sum = A.sum + B.sum;
	x.maxl = max(A.maxl , A.sum + B.maxl);
	x.maxr = max(B.maxr , B.sum + A.maxr);
	x.max = max(max(A.max,B.max),A.maxr + B.maxl);
	return x; 
} 
void build(int k,int l,int r)
{
	if(l == r)
	{
		a[k].sum = a[k].maxl = a[k].max = a[k].maxr = c[l];
		return;
	}
	int mid = (l+r) / 2;
	build(k*2,l,mid);
	build(k*2+1,mid+1,r);
	a[k] = pushup(a[k*2],a[k*2+1]);
}
void change(int k,int l,int r,int s,int t)
{
	if(l == r)
	{
		a[k].sum = a[k].maxl = a[k].max = a[k].maxr = t;
		return;
	}
	int mid = (r+l)/2;
	if(mid >= s) change(k*2,l,mid,s,t);
	else change(k*2+1,mid+1,r,s,t);
	a[k] = pushup(a[k*2],a[k*2+1]);
} 
P find(int k,int l,int r,int s,int t)
{
	P ans1,ans2;
	if(s <= l && r <= t)
	{
		return a[k];
	}
	int mid = (l+r) / 2;
	if( s <= mid && mid < t)
		return pushup(find(k*2,l,mid,s,t),find(k*2+1,mid+1,r,s,t)); 
	else if(s <= mid)
		return find(k*2,l,mid,s,t); 
	else
		return find(k*2+1,mid+1,r,s,t);
}
int main()
{
	cin >> n >> m;
	for(int i=1; i <= n ;i++)
	{
		cin >> c[i];
	}
	build(1,1,n);
	for(int i=1; i <= m; i++)
	{
		int op,x,y;
		cin >> op;
		if(op == 1)
		{
			cin >> x >> y;
			P ans = find(1,1,n,min(x,y),max(x,y)); 
			cout << ans.max << endl;
		}
		else
		{
			cin >> x >> y;
			change(1,1,n,x,y);
		}
	}
	return 0;
}

根据线段树是前序遍历的性质，写了个不用结构体但总复杂度为 $O(n\log{n} + m\log^2{n})$ 的劣解。

#include <bits/stdc++.h>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define int long long
using namespace std;
const int MAXN = 5e5 + 5;
int tr[MAXN<<2][4], a[MAXN], p;
//和，前缀，后缀，子段 0-3
int seg[MAXN];
int n, m;
void modfiy(int rt, int x)
{
	tr[rt][0] = tr[rt][1] = tr[rt][2] = tr[rt][3] = x;
}
void push_up(int rt)
{
	tr[rt][0] = tr[rt<<1][0] + tr[rt<<1|1][0];
	tr[rt][1] = max(tr[rt<<1][1], tr[rt<<1][0] + tr[rt<<1|1][1]);
	tr[rt][2] = max(tr[rt<<1|1][2], tr[rt<<1|1][0] + tr[rt<<1][2]);
	tr[rt][3] = max({tr[rt<<1][3], tr[rt<<1|1][3], tr[rt<<1][2] + tr[rt<<1|1][1]});
}
void query(int L, int R, int l, int r, int rt)
{
	if (L <= l && r <= R)
	{
		seg[++p] = rt;
		return;
	}
	int mid = (l + r) / 2;
	if (mid >= L) query(L, R, lson);
	if (mid < R) query(L, R, rson);
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	build(1, n, 1);
	for (int i = 1; i <= m; i++)
	{
		int k, l, r;
		cin >> k >> l >> r;
		if (k == 1)
		{
			p = 0;
			query(min(l, r), max(l, r), 1, n , 1);
			int ans = -(1<<30), s = 0;
			for (int i = 1; i <= p; i++)
			{
				ans = max(ans, tr[seg[i]][3]); s = 0;
				for (int j = i+1; j <= p; j++)
				{
					ans = max(ans, tr[seg[i]][2] + s + tr[seg[j]][1]);
					s += tr[seg[j]][0];
				}
			}
			cout << ans << '\n';
		}
		else update(l, r, 1, n, 1);
	}
	return 0；
}

#include<bits/stdc++.h>
#define mid (l+r>>1)
using namespace std;
#define N 500001
struct Node{int maxv,maxl,maxr,sumv;}a[N<<2];
Node pushup(Node A,Node B)
{
	Node p;
	p.maxv=max(A.maxv,B.maxv);
	p.maxv=max(p.maxv,A.maxr+B.maxl);
	p.maxl=max(A.maxl,A.sumv+B.maxl);
	p.maxr=max(B.maxr,B.sumv+A.maxr);
	p.sumv=A.sumv+B.sumv;
	return p;
}
void build(int p,int l,int r)
{
	if(l==r)
	{
	  	scanf("%d",&a[p].maxv);
	  	a[p].sumv=a[p].maxl=a[p].maxr=a[p].maxv;
	  	return;
	}
	build(p<<1,l,mid);
	build(p<<1|1,mid+1,r);
	a[p]=pushup(a[p<<1],a[p<<1|1]);
}
void update(int p,int l,int r,int s,int t)//把s改成t 
{
	if(l==r)
	{
	  	a[p].sumv=a[p].maxl=a[p].maxr=a[p].maxv=t;
	  	return;
	}
	if(s<=mid) update(p<<1,l,mid,s,t);
	else update(p<<1|1,mid+1,r,s,t);
	a[p]=pushup(a[p<<1],a[p<<1|1]);
}
Node query(int p,int l,int r,int s,int t)
{
	if(l>=s&&r<=t) return a[p];
	if(s<=mid && t>mid)
	  {
	  	Node res;
	  	res=pushup(query(p<<1,l,mid,s,t),query(p<<1|1,mid+1,r,s,t));
	  	return res;
	  }
	else if(s<=mid) return query(p<<1,l,mid,s,t);
	else return query(p<<1|1,mid+1,r,s,t);
}
int n,m;
int main()
{
	freopen("in.txt","r",stdin);
	int op,x,y;
	scanf("%d%d",&n,&m);
	build(1,1,n);
	for(int i=1;i<=m;++i){
	  	scanf("%d%d%d",&op,&x,&y);
	  	if(op==1){
		  	if(x>y) swap(x,y);
		  	printf("%d\n",query(1,1,n,x,y).maxv);
		}
	  	else update(1,1,n,x,y);
	}
	return 0;
}