Codeforces 1149 C線分ツリーLCA

題意:括弧のシーケンスをあげます.この括弧のシーケンスは二叉木を決定します.q回の問い合わせがあり、毎回この木の直径を尋ねて出力します.
考え方:https://blog.csdn.net/Huah_2018/article/details/89788074
コード:

#include 
#define ls (x << 1)
#define rs ((x << 1) | 1)
using namespace std;
const int maxn = 200010;
char s[maxn];
struct SegmentTree {
	int mx, mi, lmx, rmx, sum, ans;
	void init(int x) {
		mx = x, mi = x, lmx = -x, rmx = -x, sum = x, ans = 0;
	};
};
SegmentTree tr[maxn * 4];
void pushup(int x) {
	int k = x;
	tr[x].mx = max(tr[ls].mx, tr[rs].mx + tr[ls].sum);
	tr[x].mi = min(tr[ls].mi, tr[rs].mi + tr[ls].sum);
	tr[x].lmx = max(tr[ls].lmx, tr[rs].lmx - tr[ls].sum);
	tr[x].lmx = max(tr[x].lmx, tr[ls].mx - 2 * (tr[rs].mi + tr[ls].sum));
	tr[x].rmx = max(tr[ls].rmx, tr[rs].rmx - tr[ls].sum);
	tr[x].rmx = max(tr[x].rmx, tr[rs].mx - 2 * tr[ls].mi + tr[ls].sum);
	tr[x].sum = tr[ls].sum + tr[rs].sum;
	tr[x].ans = max(tr[ls].ans, tr[rs].ans);
	tr[x].ans = max(tr[x].ans, max(tr[ls].lmx + tr[rs].mx + tr[ls].sum, tr[rs].rmx + tr[ls].mx - tr[ls].sum));
}
void build(int x, int l, int r) {
	if(l == r) {
		tr[x].init(s[l] == '(' ? 1 : -1);
		return;
	}
	int mid = (l + r) >> 1;
	build(ls, l, mid);
	build(rs, mid + 1, r);
	pushup(x);
}
void update(int x, int l, int r, int pos) {
	if(l == r) {
		tr[x].init(s[l] == '(' ? 1 : -1);
		return;
	}
	int mid = (l + r) >> 1;
	if(pos <= mid) update(ls, l, mid, pos);
	else update(rs, mid + 1, r, pos);
	pushup(x);
}
int main() {
	int n, T;
	scanf("%d%d", &n, &T);
	n = 2 * n - 2;
	int u, v;
	scanf("%s",s + 1);
	build(1, 1, n);
	printf("%d
", tr[1].ans);
	while(T--) {
		scanf("%d%d", &u, &v);
		swap(s[u], s[v]);
		update(1, 1, n, u);
		update(1, 1, n, v);
		printf("%d
", tr[1].ans);
	}
} 

　　
転載先:https://www.cnblogs.com/pkgunboat/p/10920346.html