SPOJ - GSS1 Can you answer these queries I 线段树最大子段和

线段树求最大子段和。
对于一个区间，记录它的前缀最大子段和、后缀最大子段和、区间和、和区间最大子段和，对于每个节点，更新前缀和后缀最大子段和，则当前区间的最大子段和为左区间后缀最大值+右区间前缀最大值和左右区间最大子段和的最大值。
查询复杂度$O(logn)$

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<string.h>
#include<iostream>
#include<algorithm>

#define ll long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
const long long INF = 1e18;
const int mod = 1e9 + 7;
const int N = 5e4 + 10;
typedef std::pair<int, int> pii;
typedef std::pair<ll, ll> pll;

ll gcd(ll p, ll q) { return q == 0 ? p : gcd(q, p % q); }

using namespace std;
int n, a[N];
struct node {
    int l, r;
    ll s, v, pre, suf;
} tree[N << 2];

inline void pushup(int id) {
    int l = id << 1, r = id << 1 | 1;
    tree[id].s = tree[l].s + tree[r].s;
    tree[id].v = max(max(tree[l].v, tree[r].v), tree[l].suf + tree[r].pre);
    tree[id].pre = max(tree[l].pre, tree[l].s + tree[r].pre);
    tree[id].suf = max(tree[r].suf, tree[l].suf + tree[r].s);
}

void build(int id, int l, int r) {
    tree[id].l = l;
    tree[id].r = r;
    if (l == r) {
        tree[id].v = tree[id].s = tree[id].pre = tree[id].suf = a[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(id << 1, l, mid);
    build(id << 1 | 1, mid + 1, r);
    pushup(id);
}

node query(int id, int ql, int qr) {
    int l = tree[id].l, r = tree[id].r;
    if (ql <= l && qr >= r)
        return tree[id];
    int mid = (l + r) >> 1;
    if (qr <= mid)
        return query(id << 1, ql, qr);
    else if (ql > mid)
        return query(id << 1 | 1, ql, qr);
    else {
        node ans, ls = query(id << 1, ql, qr), rs = query(id << 1 | 1, ql, qr);
        ans.s = ls.s + rs.s;
        ans.v = max(max(ls.v, rs.v), ls.suf + rs.pre);
        ans.pre = max(ls.pre, ls.s + rs.pre);
        ans.suf = max(rs.suf, rs.s + ls.suf);
        return ans;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    build(1,1,n);
    int m, x, y;
    cin >> m;
    for (int i = 1; i <= m; i++) {
        cin >> x >> y;
        cout << query(1, x, y).v << endl;
    }
    return 0;
}