SP1716 GSS3 - Can you answer these queries III 线段树

题目传送门：SP1043 GSS1 - Can you answer these queries I

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动态维护子段和最大值

前置知识

静态维护子段和最大值：SP1043 GSS1 - Can you answer these queries I

题解传送

题解：

提供结构体指针线段树写法：

设\(l\)为区间左端点， \(r\)为区间右端点；

\(ls\)为以\(l\)为左端点的最大子段和, \(rs\)为以\(r\)为右端点的最大子段和;

\(sum\)为区间和, \(val\)为区间子段和最大和。

\(ch[0]\)为左儿子, \(ch[1]\)为右儿子.

考虑维护：

o->sum = ch[0]->sum + ch[1]->sum;
o->ls = max(ch[0]->ls, ch[0]->sum + ch[1]->ls);
o->rs = max(ch[1]->rs, ch[1]->sum + ch[0]->rs);
o->val = max(max(ch[0]->val, ch[1]->val), max(max(o->ls, o->rs), ch[0]->rs + ch[1]->ls);

这样可以考虑到所有的转移情况；

在询问时, 若询问区间跨过左右两个子区间, 则我们利用\(ls, rs, sum\)来更新出当前区间的\(val\);

所以我们返回结构体指针。

题目中为单点修改，考虑如何写up()函数。

发现维护方式与建树时相同。

void up() {
	sum = ch[0]->sum + ch[1]->sum;
	ls = Max(ch[0]->ls, ch[0]->sum + ch[1]->ls);
	rs = Max(ch[1]->rs, ch[1]->sum + ch[0]->rs);
	val = Max(Max(ch[0]->val, ch[1]->val), Max(Max(ls, rs), ch[0]->rs + ch[1]->ls));
}

直接乱搞即可。

code：

#include <iostream>
#include <cstdio>
using namespace std;
const int N = 5e4 + 5;
int read() {
	int x = 0, f = 1; char ch;
	while(! isdigit(ch = getchar())) (ch == '-') && (f = -f);
	for(x = ch^48; isdigit(ch = getchar()); x = (x<<3) + (x<<1) + (ch^48));
	return x * f;
}
int n, m;
template <class T> T Max(T a, T b) { return a > b ? a : b; }
struct Segment {
	struct node {
		int l, r, ls, rs, sum, val;
		node *ch[2];
		node(int l, int r, int ls, int rs, int sum, int val) : l(l), r(r), ls(ls), rs(rs), sum(sum), val(val) {
			ch[0] = ch[1] = NULL;
		}
		inline int mid() { return (l + r) >> 1; }
		inline void up() {
			sum = ch[0]->sum + ch[1]->sum;
			ls = Max(ch[0]->ls, ch[0]->sum + ch[1]->ls);
			rs = Max(ch[1]->rs, ch[1]->sum + ch[0]->rs);
			val = Max(Max(ch[0]->val, ch[1]->val), Max(Max(ls, rs), ch[0]->rs + ch[1]->ls));
		}
	} *root;
	void build(node *&o, int l, int r) {
		o = new node(l, r, 0, 0, 0, 0);
		if(l == r) {
			o->ls = o->rs = o->sum = o->val = read();
			return;
		}
		build(o->ch[0], l, o->mid());
		build(o->ch[1], o->mid() + 1, r);
		o->up();
	}
	void change(node *o, int x, int v) {
		if(o->l == x && o->r == x) {
			o->ls = o->rs = o->sum = o->val = v;
			return;
		}
		if(x <= o->mid()) change(o->ch[0], x, v);
		if(x > o->mid()) change(o->ch[1], x, v);
		o->up();
	}
	node *query(node *o, int l, int r) {
		if(l <= o->l && o->r <= r) return o;
		if(r <= o->mid()) return query(o->ch[0], l, r);
		if(l > o->mid()) return query(o->ch[1], l, r);
		node *res = new node(l, r, 0, 0, 0, 0);
		node *t1 = query(o->ch[0], l, r), *t2 = query(o->ch[1], l, r);
		res->sum = t1->sum + t2->sum;
		res->ls = Max(t1->ls, t1->sum + t2->ls);
		res->rs = Max(t2->rs, t2->sum + t1->rs);
		res->val = Max(Max(t1->val, t2->val), Max(Max(res->ls, res->rs), t1->rs + t2->ls));
		return res;
	}
} tr;
int main() {
	n = read();
	tr.build(tr.root, 1, n);
	m = read();
	for(int i = 1, opt, l, r; i <= m; ++ i) {
		opt = read(); l = read(); r = read();
		if(opt & 1) printf("%d\n", tr.query(tr.root, l, r)->val);
		else tr.change(tr.root, l, r);
	}
	return 0;
}