ACwing you can answer these questions do (tree line and seeking maximum continuous field)

Given the number of columns of length N A, and M instructions, each instruction may be either of the following:

1, "1 xy", the query interval [x, y] is the maximum continuous and sub-segment, i.e.

2, "2 xy", the A [x] into y.

For each query command, output a integer answer.

Input Format

The first line of two integers N, M.

The second row of the N integers A [i].

Next, 3 M lines each integers k, x, y, k = 1 indicates that the query (in this case, if x> y, swap the x, y), k = 2 represents a modification.

Output Format

For each query command output represents an integer answer.

Each answer per line.

data range

Sample input:

Sample output:

2
-1

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=5e5+5;
typedef long long ll;
using namespace std;
struct node
{
    int l,r;
    ll sum,suml,sumr,summax;
}tree[maxn<<2];
void pushup(int m)
{
    tree[m].sum=(tree[m<<1].sum+tree[m<<1|1].sum);
    tree[m].summax=max(max(tree[m<<1].summax,tree[m<<1|1].summax),tree[m<<1].sumr+tree[m<<1|1].suml);
    tree[m].suml=max(tree[m<<1].suml,tree[m<<1].sum+tree[m<<1|1].suml);
    tree[m].sumr=max(tree[m<<1|1].sumr,tree[m<<1].sumr+tree[m<<1|1].sum);
}
void build(int m,int l,int r)
{
    tree[m].l=l;
    tree[m].r=r;
    if(l==r)
    {
        scanf("%lld",&tree[m].sum);
        tree[m].summax=tree[m].sum;
        tree[m].suml=tree[m].sum;
        tree[m].sumr=tree[m].sum;
        return ;
    }
    int mid=(l+r)>>1;
    build(m<<1,l,mid);
    build(m<<1|1,mid+1,r);
    pushup(m);
}
void update(int m,int pos,ll val)
{
    if(tree[m].l==tree[m].r&&tree[m].l==pos)
    {
        tree[m].sum=val;
        tree[m].summax=tree[m].sum;
        tree[m].suml=tree[m].sum;
        tree[m].sumr=tree[m].sum;
        return ;
    }
    int mid=(tree[m].l+tree[m].r)>>1;
    if(pos<=mid)
    {
        update(m<<1,pos,val);
    }
    else
    {
        update(m<<1|1,pos,val);
    }
    pushup(m);
}
node query(int m,int l,int r)
{
    if(tree[m].l==l&&tree[m].r==r)
    {
        return tree[m];
    }
    int mid=(tree[m].l+tree[m].r)>>1;
    if(r<=mid)
    {
        return query(m<<1,l,r);
    }
    else if(l>mid)
    {
        return query(m<<1|1,l,r);
    }
    else
    {
        node treel,treer,temp;
        treel=query(m<<1,l,mid);
        treer=query(m<<1|1,mid+1,r);
        temp.sum=treel.sum+treer.sum;
        temp.summax=max(max(treel.summax,treer.summax),treel.sumr+treer.suml);
        temp.suml=max(treel.suml,treel.sum+treer.suml);
        temp.sumr=max(treer.sumr,treer.sum+treel.sumr);
        return temp;
    }
}
int main()
{
    
    int n,q;
    cin>>n>>q;
    build(1,1,n);
    int op,l,r;
    while(q--)
    {
        scanf("%d",&op);
        if(op==1)
        {
            scanf("%d%d",&l,&r);
            if(l>r)
            {
                swap(l,r);
            }
            printf("%lld\n",query(1,l,r).summax);
        }
        else
        {
            scanf("%d%d",&l,&r);
    
            update(1,l,r);
        }
    }
    return 0;
}