SPOJ GSS3 Can you answer these queries III（线段树）

Can you answer these queries III

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

解题思路：

在询问一段数列的连续最大自序列和的基础上，增加了单点修改。

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 50005;
int a[N];
struct node{
    int l,r,lx,rx,mx,sum;
}tree[N<<2];

void pushUp(int id){
    tree[id].lx = max(tree[id<<1].lx,tree[id<<1|1].lx+tree[id<<1].sum);
    tree[id].rx = max(tree[id<<1|1].rx,tree[id<<1].rx+tree[id<<1|1].sum);
    tree[id].mx = max(max(tree[id<<1].mx,tree[id<<1|1].mx),tree[id<<1].rx+tree[id<<1|1].lx);
    tree[id].sum = tree[id<<1].sum+tree[id<<1|1].sum;
}

void build(int id,int l,int r){
    tree[id].l = l;
    tree[id].r = r;
    if(l == r){
        tree[id].lx = tree[id].rx = tree[id].mx = tree[id].sum = a[l];
        return ;
    }
    int mid = (l+r)/2;
    build(id<<1,l,mid);
    build(id<<1|1,mid+1,r);
    pushUp(id);
}
int Q_L(int id,int l,int r){
    if(tree[id].l == l && tree[id].r == r)
        return tree[id].lx;
    int mid = (tree[id].l+tree[id].r)/2;
    if(r <= mid)
        return Q_L(id<<1,l,r);
    else if(l > mid)
        return Q_L(id<<1|1,l,r);
    else return max(Q_L(id<<1,l,mid),max(tree[id<<1].sum,tree[id<<1].sum+Q_L(id<<1|1,mid+1,r)));
}
int Q_R(int id,int l,int r){
    if(tree[id].l == l && tree[id].r == r)
        return tree[id].rx;
    int mid = (tree[id].l+tree[id].r)/2;
    if(r <= mid)
        return Q_R(id<<1,l,r);
    else if(l > mid)
        return Q_R(id<<1|1,l,r);
    else return max(Q_R(id<<1|1,mid+1,r),max(tree[id<<1|1].sum,tree[id<<1|1].sum+Q_R(id<<1,l,mid)));
}

void update(int id,int x,int val){
    if(tree[id].l == tree[id].r){
        tree[id].mx += val;
        tree[id].rx += val;
        tree[id].lx += val;
        tree[id].sum += val;
        return ;
    }
    int mid = (tree[id].l+tree[id].r)/2;
    if(x <= mid)
        update(id<<1,x,val);
    else
        update(id<<1|1,x,val);
    pushUp(id);
}

int query(int id,int l,int r){
    if(tree[id].l == l && tree[id].r == r)
        return tree[id].mx;
    int mid = (tree[id].l+tree[id].r)/2;
    if(r <= mid)
        return query(id<<1,l,r);
    else if(l > mid)
        return query(id<<1|1,l,r);
    else return max(max(query(id<<1,l,mid),query(id<<1|1,mid+1,r)),Q_R(id<<1,l,mid)+Q_L(id<<1|1,mid+1,r));
}

int main(){
    int n,m;
    while(~scanf("%d",&n)){
        for(int i = 1; i <= n; i++)
            scanf("%d",&a[i]);
        build(1,1,n);
        int op,x,y;
        scanf("%d",&m);
        while(m--){
            scanf("%d%d%d",&op,&x,&y);
            if(op == 0){
                update(1,x,y-a[x]);
                a[x] = y;
            }
            else
                printf("%d\n",query(1,x,y));
        }
    }
    return 0;
}