标签：序列 tail href pos turn log 树的直径 max tree
题意：给你一个括号序列，这个括号序列将确定一颗二叉树。有q次询问，每次询问输出这颗树的直径。
思路：https://blog.csdn.net/Huah_2018/article/details/89788074
代码：
#include <bits/stdc++.h>
#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\n", 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\n", tr[1].ans);
}
}
标签：序列 tail href pos turn log 树的直径 max tree
原文地址：https://www.cnblogs.com/pkgunboat/p/10920346.html
