做题记录 Luogu P4513

Posted 2021-06-20

Luogu P4513 小白逛公园

线段树维护区间最大子列模板（只用单点修改，还要弱一点）

今天终于理解了 root << 1 和 tree[root].l 的区别了：前者是 root 号节点的左子节点，后者是 root 号节点所代表的区间左下标

#include<bits/stdc++.h>
using namespace std;
#define maxn 500005
struct node
{
	int l, r, val, mxqz, mxhz, mxzd;
};
node tree[maxn * 4];
int num[maxn], n, m;
void pushup(int root)
{
	tree[root].val = tree[root << 1].val + tree[root << 1 | 1].val;
	tree[root].mxqz = max(tree[root << 1].mxqz, tree[root << 1].val + tree[root << 1 | 1].mxqz);
	tree[root].mxhz = max(tree[root << 1 | 1].mxhz, tree[root << 1 | 1].val + tree[root << 1].mxhz);
	tree[root].mxzd = max(tree[root << 1].mxzd, tree[root << 1 | 1].mxzd);
	tree[root].mxzd = max(tree[root].mxzd, tree[root << 1].mxhz + tree[root << 1 | 1].mxqz);
	return;
}
void build(int root, int l, int r)
{
	tree[root].l = l;
	tree[root].r = r;
	if(l == r)
	{
		tree[root].val = tree[root].mxqz = tree[root].mxhz = tree[root].mxzd = num[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(root << 1, l, mid);
	build(root << 1 | 1, mid + 1, r);
	pushup(root);
	return;
}
void update(int root, int pos, int x)
{
	if(tree[root].l == tree[root].r)
	{
		tree[root].val = tree[root].mxqz = tree[root].mxhz = tree[root].mxzd = x;
		return;
	}
	int mid = (tree[root].l + tree[root].r) >> 1;
	if(pos <= mid)
	{
		update(root << 1, pos, x);
	}
	else
	{
		update(root << 1 | 1, pos, x);
	}
	pushup(root);
	return;
}
node query(int root, int L, int R)
{
	if(L <= tree[root].l && R >= tree[root].r)
	{
		return tree[root];
	}
	int mid = (tree[root].l + tree[root].r) >> 1;
	if(R <= mid)
	{
		return query(root << 1, L, R);
	}
	if(L > mid)
	{
		return query(root << 1 | 1, L, R);
	}
	node lch = query(root << 1, L, R), rch= query(root << 1 | 1, L, R), ret;
	ret.val = lch.val + rch.val;
	ret.mxqz = max(lch.mxqz, lch.val + rch.mxqz);
	ret.mxhz = max(rch.mxhz, rch.val + lch.mxhz);
	ret.mxzd = max(lch.mxzd, rch.mxzd);
	ret.mxzd = max(ret.mxzd, lch.mxhz + rch.mxqz);
	return ret;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for(int i = 1; i <= n; i++)
	{
		cin >> num[i];
	}
	build(1, 1, n);
	int opt, x, y;
	for(int i = 1; i <= m; i++)
	{
		cin >> opt;
		if(opt == 1)
		{
			cin >> x >> y;
			if(x > y)
			{
				swap(x, y);
			}
			node ans = query(1, x, y);
			cout << ans.mxzd << endl;
		}
		else if(opt == 2)
		{
			cin >> x >> y;
			update(1, x, y);
		}
	}
	return 0;
}