SPOJ GSS3 Can you answer these queries III[线段树]

Description

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.

Example

Input:
4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

Output:
6
4
-3

GSS1加个单点修改

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define m ((l+r)>>1)
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc o<<1
#define rc o<<1|1
using namespace std;
typedef long long ll;
const int N=5e5+5,INF=2e9+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,q,op,x,y;
struct node{
    int sum,mx,pre,suf;
}t[N<<2];
void merge(int o){
    t[o].sum=t[lc].sum+t[rc].sum;
    t[o].mx=max(t[lc].suf+t[rc].pre,max(t[lc].mx,t[rc].mx));
    t[o].pre=max(t[lc].pre,t[lc].sum+t[rc].pre);
    t[o].suf=max(t[rc].suf,t[rc].sum+t[lc].suf);
}
void build(int o,int l,int r){
    if(l==r) t[o].sum=t[o].mx=t[o].pre=t[o].suf=read();
    else{
        build(lson);
        build(rson);
        merge(o);
    }
}
int qpre(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o].pre;
    else if(qr<=m) return qpre(lson,ql,qr);
    else return max(qpre(lson,ql,qr),t[lc].sum+qpre(rson,ql,qr));
}
int qsuf(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o].suf;
    else if(m<ql) return qsuf(rson,ql,qr);
    else return max(t[rc].suf,t[rc].sum+qsuf(lson,ql,qr));
}
int qmx(int o,int l,int r,int ql,int qr){
    if(ql<=l&&r<=qr) return t[o].mx;
    else{
        int ans=-INF;
        if(ql<=m) ans=max(ans,qmx(lson,ql,qr));
        if(m<qr) ans=max(ans,qmx(rson,ql,qr));
        if(ql<=m&&m<qr) ans=max(ans,qsuf(lson,ql,qr)+qpre(rson,ql,qr));
        return ans;
    }
}
void update(int o,int l,int r,int p,int v){
    if(l==r) t[o].sum=t[o].mx=t[o].pre=t[o].suf=v;
    else{
        if(p<=m) update(lson,p,v);
        else update(rson,p,v);
        merge(o);
    }
}
int main(){
    n=read();
    build(1,1,n);
    q=read();
    for(int i=1;i<=q;i++){
        op=read();x=read();y=read();
        if(op) printf("%d
",qmx(1,1,n,x,y));
        else update(1,1,n,x,y);
    }
}