Codeforces 1150E（树、线段树）

要点

• 括号序列平衡度即树深度的性质
• 相当于中序遍历，则两点间最浅的地方即是LCA的性质
• 线段树维护\(d(a) + d(c) - 2*d(lca(a,c))\)，一层层剥，思考维护这个量需要什么，结果维护一大堆。

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 2e5 + 5;

int n, q;
char s[maxn];

class SegmentTree {
public:
	#define ls (p << 1)
	#define rs (p << 1 | 1)

	struct node {
		int l, r;
		int depth;//its root is s[l]
		int ans;//d(a) + d(c) - 2d(lca(a, c))
		int lmax;//d(a) - 2d(lca(a, c))
		int rmax;//d(c) - 2d(lca(a, c))
		int mxd;//max depth
		int mnd;//min depth

		void init(int val) {
			mxd = mnd = depth = val;
			lmax = rmax = -val;
			ans = 0;
		}
	}t[maxn * 3];

	void Push_up(int p) {
		// b = lca(a, c), a <= b <= c
		t[p].depth = t[ls].depth + t[rs].depth;
		t[p].mxd = max(t[ls].mxd, t[rs].mxd + t[ls].depth);
		t[p].mnd = min(t[ls].mnd, t[rs].mnd + t[ls].depth);
		t[p].lmax = max(t[ls].lmax, t[rs].lmax - t[ls].depth);//a, b both in l or r
		t[p].lmax = max(t[p].lmax, t[ls].mxd - 2 * (t[rs].mnd + t[ls].depth));//a in l and b in r
		t[p].rmax = max(t[ls].rmax, t[rs].rmax - t[ls].depth);//b, c both in l or r
		t[p].rmax = max(t[p].rmax, t[rs].mxd + t[ls].depth - 2 * t[ls].mnd);//b int l and c in r
		t[p].ans = max(t[ls].ans, t[rs].ans);//a,b,c all in l or r
		t[p].ans = max(t[p].ans, max(t[ls].lmax + t[rs].mxd + t[ls].depth, t[ls].mxd + t[rs].rmax - t[ls].depth));//ab in l, c in r || a in l, bc in r
	}

	void Build(int l, int r, int p) {
		t[p].l = l, t[p].r = r;
		if (l == r) {
			t[p].init(s[l] == '(' ? 1 : -1);
			return;
		}
		int mid = (l + r) >> 1;
		Build(l, mid, ls);
		Build(mid + 1, r, rs);
		Push_up(p);
	}

	void Modify(int l, int r, int p) {
		if (t[p].l == t[p].r) {
			t[p].init(s[l] == '(' ? 1 : -1);
			return;
		}
		int mid = (t[p].l + t[p].r) >> 1;
		if (l <= mid)	Modify(l, r, ls);
		if (mid < r)	Modify(l, r, rs);
		Push_up(p);
	}
}tree;

int main() {
	scanf("%d %d", &n, &q);
	scanf("%s", s + 1);

	n = (n - 1) << 1;
	tree.Build(1, n, 1);
	printf("%d\n", tree.t[1].ans);

	for (int a, b; q; q--) {
		scanf("%d %d", &a, &b);
		if (s[a] != s[b]) {
			swap(s[a], s[b]);
			tree.Modify(a, a, 1);
			tree.Modify(b, b, 1);
		}
		printf("%d\n", tree.t[1].ans);
	}
}