CH4301 Can you answer on these queries III 题解

Given the number of columns of length N A, and M instructions (N≤500000, M≤100000), each instruction may be either of the following:
"2 XY", the A [x] into y.
"1 xy", the query interval [x, y] is the maximum continuous and sub-segment, i.e., max (x≤l≤r≤y) ⁡ { \ (\ sum_ {I} = L ^ R & lt \) A [I] }.
For each inquiry, output a integer answer.

A template title, interval tree line and the largest sub-segment.

Code

#include<bits/stdc++.h>
using namespace std;
const int SIZE=500010;
struct node{
    long long data;
    long long sum;
    long long l,r;
    long long lmax,rmax;
}t[SIZE*4];
long long n,m,u,v,k,sum,a[SIZE];
void build(long long p,long long l,long long r){
    t[p].l=l,t[p].r=r;
    if(l==r){
        t[p].sum=t[p].data=t[p].lmax=t[p].rmax=a[l];
        return;
    } 
    long long mid=(l+r)/2;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    t[p].sum=t[p*2].sum+t[p*2+1].sum;
    t[p].lmax=max(t[p*2].lmax,t[p*2].sum+t[p*2+1].lmax);
    t[p].rmax=max(t[p*2+1].rmax,t[p*2+1].sum+t[p*2].rmax);
    t[p].data=max(max(t[p*2].data,t[p*2+1].data),t[p*2].rmax+t[p*2+1].lmax);
}
void change(long long p,long long x,long long v){
    if(t[p].l==t[p].r){
        t[p].data=t[p].sum=t[p].lmax=t[p].rmax=v;
        return;
    }
    long long mid=(t[p].l+t[p].r)/2;
    if(x<=mid) change(p*2,x,v);
    else change(p*2+1,x,v);
    t[p].sum=t[p*2].sum+t[p*2+1].sum;
    t[p].lmax=max(t[p*2].lmax,t[p*2].sum+t[p*2+1].lmax);
    t[p].rmax=max(t[p*2+1].rmax,t[p*2+1].sum+t[p*2].rmax);
    t[p].data=max(max(t[p*2].data,t[p*2+1].data),t[p*2].rmax+t[p*2+1].lmax);    
}
node ask(long long p,long long l,long long r){
    if(l<=t[p].l&&r>=t[p].r) return t[p];
    long long mid=(t[p].l+t[p].r)/2;
    if(mid>=r) return ask(p*2,l,r);
    if(mid<l) return ask(p*2+1,l,r);
    else{
        node ans,a1,b;
        a1=ask(p*2,l,r);
        b=ask(p*2+1,l,r);
        ans.sum=a1.sum+b.sum;
        ans.data=max(max(a1.data,a1.rmax+b.lmax),b.data);
        ans.lmax=max(a1.lmax,a1.sum+b.lmax);
        ans.rmax=max(b.rmax,b.sum+a1.rmax);
        return ans;
    }
}
int main(){
    scanf("%lld",&n);
    scanf("%lld",&m);
    for(int i=1;i<=n;++i){
        scanf("%lld",&a[i]);
    } 
    build(1,1,n);
    while(m--){
        scanf("%lld %lld %lld",&k,&u,&v);
        if(k==1){
            if(u>v) swap(u,v);
            printf("%lld\n",ask(1,u,v).data);
        }
        else{
            change(1,u,v);
        }
    }
    return 0;
}