AcWing 245. 你能回答这些问题吗

题目链接：传送门

线段树单点修改+查询最大子段和
一边写对？
只不过忘了改数组大小没看到$x>y$而已
维护区间和$w$，区间最大子段和$f$，从左端点开始的最大子段和$lf$，从右端点开始的最大子段和$rf$
$w$照样维护，$f$取左儿子和右儿子的$max$，还要和左儿子的$rf+$右儿子的$lf$取$max$，$lf$对左儿子的$lf$ 和 左儿子的$w+$右儿子的$lf$取$max$，$rf$对右儿子的$rf$ 和 右儿子的$w+$右儿子的$rf$取$max$，这样简单维护起来就好，查询的时候一样

#include 
#define A 2000010

using namespace std;
typedef long long ll;
struct node {int l, r, w, f, lf, rf;}tree[A];
int n, m, opt, a, b;
void up(int k) {
	tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w;
	tree[k].f = max(tree[k << 1].f, tree[k << 1 | 1].f);
	tree[k].f = max(tree[k].f, tree[k << 1].rf + tree[k << 1 | 1].lf);
	tree[k].lf = max(tree[k << 1].lf, tree[k << 1].w + tree[k << 1 | 1].lf);
	tree[k].rf = max(tree[k << 1 | 1].rf, tree[k << 1 | 1].w + tree[k << 1].rf);
}
void build(int k, int l, int r) {
	tree[k].l = l; tree[k].r = r;
	if (l == r) {
		int d; scanf("%d", &d);
		tree[k].w = tree[k].f = tree[k].lf = tree[k].rf = d;
		return;
	}
	int m = (l + r) >> 1;
	build(k << 1, l, m); build(k << 1 | 1, m + 1, r);
	up(k);
}
void change(int k, int pos, int val) {
	if (tree[k].l == tree[k].r) {
		tree[k].w = tree[k].f = tree[k].lf = tree[k].rf = val;
		return;
	}
	int m = (tree[k].l + tree[k].r) >> 1;
	if (pos <= m) change(k << 1, pos, val);
	else change(k << 1 | 1, pos, val);
	up(k);
}
node ask(int k, int l, int r) {
	if (tree[k].l >= l and tree[k].r <= r) return tree[k];
	int m = (tree[k].l + tree[k].r) >> 1;
	if (l <= m and r > m) {
		node lx = ask(k << 1, l, r), rx = ask(k << 1 | 1, l, r), tmp;
		tmp.f = max(lx.f, rx.f); tmp.f = max(tmp.f, lx.rf + rx.lf);
		tmp.lf = max(lx.lf, lx.w + rx.lf); tmp.rf = max(rx.rf, rx.w + lx.rf);
		return tmp;
	}
	if (l <= m) return ask(k << 1, l, r);
	if (r > m) return ask(k << 1 | 1, l, r);
}

int main(int argc, char const *argv[]) {
	cin >> n >> m; build(1, 1, n);
	while (m--) {
		scanf("%d%d%d", &opt, &a, &b);
		if (opt == 1) printf("%d\n", ask(1, min(a, b), max(a, b)).f);
		else change(1, a, b);
	}
	return 0;
}