GSS3 - Can you answer these queries III

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 }.

Input

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 }.

Output

For each query, print an integer as the problem required.

#include<bits/stdc++.h>
#define MAXN 50010
using namespace std;
int lc[4*MAXN],rc[4*MAXN],mc[4*MAXN],sum[4*MAXN];
int c[MAXN];
void build(int id,int l,int r)
{
	if(l==r)
	{
		sum[id]=mc[id]=lc[id]=rc[id]=c[l];
		return;
	}
	int mid=(l+r)/2;
	build(id*2,l,mid);
	build(id*2+1,mid+1,r);
	sum[id]=sum[id*2]+sum[id*2+1];
	lc[id]=max(lc[id*2],sum[id*2]+lc[id*2+1]);
	rc[id]=max(rc[id*2+1],sum[id*2+1]+rc[id*2]);
	mc[id]=max(max(mc[id*2],mc[id*2+1]),rc[id*2]+lc[id*2+1]);
}
int query(int id,int x,int y,int l,int r,int flag,int &ans)
{
	if(l<=x&&y<=r)
	{
		ans=max(ans,mc[id]);
		return flag==-1?lc[id]:rc[id];
	}
	int mid=(x+y)/2;
	if(r<=mid)
	return query(id*2,x,mid,l,r,-1,ans);
	else if(l>mid)
	return query(id*2+1,mid+1,y,l,r,1,ans);
	else
	{
		int ln=query(id*2,x,mid,l,r,1,ans);
		int rn=query(id*2+1,mid+1,y,l,r,-1,ans);
		ans=max(ln+rn,ans);
		if(flag==-1)
		return max(lc[id*2],sum[id*2]+rn);
		else
		return max(rc[id*2+1],sum[id*2+1]+ln);
	}
}
void update(int id,int l,int r,int pos,int val)
{
	if(l==r)
	{
		sum[id]=lc[id]=rc[id]=mc[id]=val;
		return;
	}
	int mid=(l+r)/2;
	if(pos<=mid)
	update(id*2,l,mid,pos,val);
	else
	update(id*2+1,mid+1,r,pos,val);
	sum[id]=sum[id*2]+sum[id*2+1];
	lc[id]=max(lc[id*2],sum[id*2]+lc[id*2+1]);
	rc[id]=max(rc[id*2+1],sum[id*2+1]+rc[id*2]);
	mc[id]=max(max(mc[id*2],mc[id*2+1]),rc[id*2]+lc[id*2+1]);
}
int main()
{
	int n;
	scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&c[i]);
    }
   	build(1,1,n);
   	int m;
   	scanf("%d",&m);
   	for(int i=0;i<m;i++)
   	{
   		int op,l,r;
   		scanf("%d%d%d",&op,&l,&r);
   		if(op==1)
   		{
   			
   			int ans=c[l];
   			query(1,1,n,l,r,-1,ans);
   			printf("%d\n",ans);
		}
		else
		{
			c[l]=r;
			update(1,1,n,l,r);
		}
	}
    return 0;
}

在GSS1的基礎上加入修改操作就行了；