#include <bits/stdc++.h>
using namespace std;
#define mid ((l + r) >> 1)
const int maxn = 5e5 + 10;
int n, m;
struct tree{
	int Amx, Bmn, lx, rx, mx;
}a[maxn << 3];
struct node {
	int a, b;
}b[maxn];
tree update(tree A, tree B) {
	tree T;
	T.Amx = max(A.Amx, B.Amx);
	T.Bmn = min(A.Bmn, B.Bmn);
	T.lx = A.Amx - B.Bmn; //需要满足A的位置在B前面
	T.rx = B.Amx - A.Bmn; //需要满足B的位置在A前面
	T.lx = max(T.lx, max(A.lx, B.lx));
	T.rx = max(T.rx, max(A.rx, B.rx));
	T.mx = max(A.lx + B.Amx, B.rx + A.Amx); //对应以上说的 3 4 操作
	T.mx = max(T.mx, max(A.mx, B.mx)); //1 2 
	return T;
}
void build(int p, int l, int r) {
	if(l == r) {
		a[p].Amx = b[l].a, a[p].Bmn = b[l].b;
		a[p].lx = a[p].rx = a[p].mx = -0x3f3f3f3f; //不保证A都大于B 所以要初始化
		return;
	}
	build(p<<1,l,mid), build(p<<1|1,mid+1,r);
	a[p] = update(a[p<<1], a[p<<1|1]);
}
void change(int p, int l, int r, int pos, int k, int op) { //单点修改
	if(l == r) {
		if(op == 1) a[p].Amx = k;
		if(op == 2) a[p].Bmn = k;
		return;
	}
	if(pos <= mid) change(p<<1,l,mid,pos,k,op);
	if(pos >  mid) change(p<<1|1,mid+1,r,pos,k,op);
	a[p] = update(a[p<<1], a[p<<1|1]);
}
tree query(int p, int l, int r, int L, int R) { //区间查询
	if(L <= l && r <= R) return a[p];
	if(L >  mid) return query(p<<1|1,mid+1,r, L, R); //全在右区间
	if(R <= mid) return query(p<<1,l,mid, L, R); //全在左区间
	return update(query(p<<1,l,mid, L, R), query(p<<1|1,mid+1,r, L, R)); 
	//两个区间合并与change build函数中一样
}
int main() {
//	freopen("in.txt","r",stdin);
	scanf("%d%d", &n, &m);
	for(int i = 1;i <= n;i++) scanf("%d", &b[i].a);
	for(int i = 1;i <= n;i++) scanf("%d", &b[i].b);
	build(1, 1, n);
	for(int i = 1;i <= m;i++) {
		int op, x, y;
		scanf("%d%d%d", &op, &x, &y);
		if(op == 1)	change(1, 1, n, x, y, 1);
		if(op == 2) change(1, 1, n, x, y, 2);
		if(op == 3) printf("%d\n", query(1, 1, n, x, y).mx);
	}
	return 0;
}

题目难度属于半裸题，难点在于下面题解看不懂 下面我浅浅说一下 其余不多说，易错点是查找的时候需要注意等号要不然会爆 言归正传 只说一下push_up（强烈建议先做前一题） 给定五个维护量 第一个max 含义Ai +Aj - Bk 最大值 Amax 含义在l-r区间内A最大值 Bmin 含义在l-r区间内B最小值 ab 含义Ai - Bk 最大值 bc 含义Aj - Bk 最大值 具体维护看代码，比较好理解 暑假的救赎day4 启动！

#include <bits/stdc++.h>
using namespace std;
struct Node{
	int Amax,Bmin,ab,bc,max;
};
Node a[4000005];
int n,m,c1[2000005],c2[2000005];

Node push_up(Node x,Node y)
{
	Node p;
	p.Amax = max(x.Amax,y.Amax);
	p.Bmin = min(x.Bmin,y.Bmin);
	p.max = max(x.Amax + y.bc , x.ab + y.Amax);
	p.max = max(max(x.max,y.max),p.max);
	p.ab = max(max(x.ab,y.ab),x.Amax - y.Bmin);
	p.bc = max(max(x.bc,y.bc),y.Amax - x.Bmin);
	return p;
}
void out(Node P)
{
	cout << P.max << " " << P.Amax << " " << P.Bmin << " " << P.ab  << " " << P.bc << endl; 
}
void build(int k,int l,int r)
{
	if(l == r)
	{
		a[k].Amax = c1[l];
		a[k].Bmin = c2[l];
		a[k].max = a[k].ab = a[k].bc = -0x3f3f3f3f;
		return;
	}
	int mid = (l+r) / 2;
	build(2*k,l,mid);
	build(2*k+1,mid+1,r);
	a[k] = push_up(a[k*2] ,a[k*2+1]);
}
void change(int k,int l,int r , int s,int t,int op)
{
	if(l == r)
	{
		if(op == 1) 
		    a[k].Amax = t;
		if(op == 2)
		    a[k].Bmin = t;
		return;
	}
	int mid = (l + r)/2;
	if(s <= mid) change(k*2,l,mid,s,t,op);
	else change(k * 2 + 1,mid+1,r,s,t,op);
	a[k] = push_up(a[k*2],a[k*2+1]); 
}
Node sum(int k,int l,int r,int s,int t)
{
	if(s <= l && r <= t)
	{
		return a[k];
	}
	int mid = (l+r) / 2;
	if(s <= mid && mid < t)
		return push_up(sum(k*2,l,mid,s,t),sum(k*2+1,mid+1,r,s,t));
	else if(s <= mid)
		return sum(k*2,l,mid,s,t);
	else
	    return sum(k*2+1,mid+1,r,s,t);
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	cin >> n >> m;
	for(int i=1; i <= n; i++)
		cin >> c1[i];
	for(int i=1; i <= n ;i++)
	    cin >> c2[i];
	build(1,1,n);
	for(int i=1; i <= m ;i++)
	{
		int op, x,y;
		cin >> op >> x >> y;
		if(op == 1)
			change(1,1,n,x,y,op);
		if(op==2)
		    change(1,1,n,x,y,op);
		if(op == 3)
		{
			Node ans = sum(1,1,n,x,y);
			cout << ans.max << '\n';
		}
	}
	return 0;
}