SP1716 GSS3 Can you answer these queries III 线段树维护最大连续子段和

https://www.luogu.org/problem/SP1716
题意翻译

nnn 个数，qqq 次操作

操作0 x y把AxA_xAx​ 修改为yyy

操作1 l r询问区间[l,r][l, r][l,r] 的最大子段和

感谢 @Edgration 提供的翻译
题目描述

You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations:
modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + … + Aj | x<=i<=j<=y }.
输入格式

The first line of input contains an integer N. The following line contains N integers, representing the sequence A1…AN.
The third line contains an integer M. The next M lines contain the operations in following form:
0 x y: modify Ax into y (|y|<=10000).
1 x y: print max{Ai + Ai+1 + … + Aj | x<=i<=j<=y }.
输出格式

For each query, print an integer as the problem required.
输入输出样例
输入 #1

4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

输出 #1

6
4
-3

思路： sum表示区间和，begl表示从区间左端点开始的最大连续子段和，begr表示从区间右端点开始的最大连续子段和，ans表示该区间的最大连续子段和。在合并左右区间时的操作：$$root.sum=lc.sum+rc.sum$$$$root.begl=max(lc.begl,lc.sum+rc.begl)$$$$root.begr=max(rc.begr,rc.sum+lc.begr)$$$$root.ans=max(lc.ans,rc.ans,lc.begr+rc.begl)$$
维护这些值就完事儿了。

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int maxn=5e4+5;

struct node
{
    int l,r,sum,begl,begr,ans;
}tree[maxn<<2];

int n,m;

inline void up(int i)
{
    int l=i<<1,r=i<<1|1;
    tree[i].sum=tree[l].sum+tree[r].sum;
    tree[i].begl=max(tree[l].begl,tree[l].sum+tree[r].begl);
    tree[i].begr=max(tree[r].begr,tree[r].sum+tree[l].begr);
    tree[i].ans=max(tree[l].begr+tree[r].begl,max(tree[l].ans,tree[r].ans));
}

inline void build(int i,int l,int r)
{
    tree[i].l=l,tree[i].r=r;
    if(l==r)
    {
        scanf("%d",&tree[i].sum);
        tree[i].ans=tree[i].begl=tree[i].begr=tree[i].sum;
        return ;
    }
    int mid=(l+r)>>1;
    build(i<<1,l,mid);
    build(i<<1|1,mid+1,r);
    up(i);
}

void update(int i,int p,int v)
{
    if(tree[i].l==tree[i].r)
    {
        tree[i].ans=tree[i].sum=tree[i].begl=tree[i].begr=v;
        return ;
    }
    int mid=(tree[i].l+tree[i].r)>>1;
    if(p<=mid)
        update(i<<1,p,v);
    else
        update(i<<1|1,p,v);
    up(i);
}

node query(int i,int l,int r)
{
    if(l==tree[i].l&&r==tree[i].r)
        return tree[i];
    int mid=(tree[i].l+tree[i].r)>>1;
    if(r<=mid)
        return query(i<<1,l,r);
    else if(l>mid)
        return query(i<<1|1,l,r);
    else
    {
        node lc=query(i<<1,l,mid);
        node rc=query(i<<1|1,mid+1,r);
        node tmp;
        tmp.sum=lc.sum+rc.sum;
        tmp.begl=max(lc.begl,lc.sum+rc.begl);
        tmp.begr=max(rc.begr,rc.sum+lc.begr);
        tmp.ans=max(lc.begr+rc.begl,max(lc.ans,rc.ans));
        return tmp;
    }
}

inline void prework()
{
    scanf("%d",&n);
    build(1,1,n);
}

inline void mainwork()
{
    scanf("%d",&m);
    int op,u,v;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&op,&u,&v);
        if(!op)
            update(1,u,v);
        else
            printf("%d\n",query(1,u,v).ans);
    }
}

int main()
{
    prework();
    mainwork();
    return 0;
}