线段树

重点
1. 根据题目分析，节点需要添加的新的信息，新信息如何维护，继续思考，直到全部信息可知
2. query的时候，需要分情况，因为查询的东西比较特殊，需要利用返回值

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 500010, M = 100010;
int a[N];
int n, m;
struct Node {
    int l, r;
    int sum, lmax, rmax, tmax;
} tr[N * 4];

void pushup(Node &u, Node &l, Node &r){
    u.sum = l.sum + r.sum;
    u.lmax = max(l.lmax, l.sum + r.lmax);
    u.rmax = max(r.rmax, r.sum + l.rmax);
    u.tmax = max(max(l.tmax, r.tmax), l.rmax + r.lmax);
}

void pushup(int u) {
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, a[l], a[l], a[l], a[l]};
    else {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

void modify(int u, int x, int v) {
    if (tr[u].l == x && tr[u].r == x) tr[u] = {x, x, v, v, v, v};
    else {
        int mid = tr[u].l + tr[u].r >> 1;
        if (x <= mid) modify(u << 1, x, v);
        else modify(u << 1 | 1, x, v);
        pushup(u);
    }
}

Node query(int u, int l, int r) {
    if (tr[u].l >= l && tr[u].r <= r) return tr[u];
    else {
        int mid = tr[u].l + tr[u].r >> 1;
        if (r <= mid) return query(u << 1, l, r);
        else if (l > mid) return query(u << 1 | 1, l, r);
        else {
            auto lnode = query(u << 1, l, r);
            auto rnode = query(u << 1 | 1, l, r);
            Node res;
            pushup(res, lnode, rnode);
            return res;
        }
    }
}

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
    }
    build(1, 1, n);

    int x, y, k;
    while (m--) {
        scanf("%d%d%d", &k, &x, &y);
        if (k == 1) {
            if (x > y) swap(x, y);
            auto t = query(1, x, y);
            printf("%d\n", t.tmax);
        } else {
            modify(1, x, y);
        }
    }
    return 0;
}