CodeForces 1311E Construct the Binary Tree
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题意
给定(n)和(d),构造一颗(n)个节点的二叉树(以(1)为根),所有节点到(1)的距离和为(d),不行输出(NO),否则输出(YES)和(2)-(n)的父亲编号
分析
最大和显然是一条链,如果最大和仍小于(d),则不行,否则先构造出一条链,然后枚举当前的层数,如果当前层数的下一层仍然可以填,则将链的尾端放置在下一层可得到最大的变化,如果这个变化大于所需要的变化,则找到挂置的层数,否则则挂在下一层,若下一层已满,则将所枚举层数下移
#pragma GCC optimize(3, "Ofast", "inline")
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
#define start ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define ll long long
#define int ll
#define ls st<<1
#define rs st<<1|1
#define pii pair<int,int>
#define rep(z, x, y) for(int z=x;z<=y;++z)
#define com bool operator<(const node &b)
const int maxn = (ll) 5e3 + 5;
const int mod = 998244353;
const int inf = 0x3f3f3f3f;
int T = 1;
int pre[maxn];
set<int> lay[maxn];
vector<int> son[maxn];
void solve() {
int n, d;
cin >> n >> d;
int now = 0;
son[1].clear();
lay[1].clear();
lay[1].insert(1);
rep(i, 2, n) {
pre[i] = i - 1, now += i - 1, lay[i].clear();
lay[i].insert(i);
son[i].clear();
son[i - 1].push_back(i);
}
if (now < d) {
cout << "NO
";
return;
}
int nowLay = 1;
for (int i = n; i >= 1; --i) {
if (i <= nowLay + 1)
break;
int maxx = i - nowLay - 1;
if (maxx >= now - d) {
if (lay[i - 1 - (now - d)].empty()) {
++nowLay;
++i;
continue;
}
int fa = *lay[i - 1 - (now - d)].begin();
pre[i] = fa;
now = d;
break;
} else {
now -= maxx;
int fa = *lay[nowLay].begin();
son[fa].push_back(i);
if (son[fa].size() == 2)
lay[nowLay].erase(fa);
son[pre[i]].erase(son[pre[i]].begin());
pre[i] = fa;
lay[nowLay + 1].insert(i);
lay[i].erase(i);
if (lay[nowLay].empty())
++nowLay;
}
}
if (now != d) {
cout << "NO
";
return;
}
cout << "YES
";
rep(i, 2, n)cout << pre[i] << ‘ ‘;
cout << ‘
‘;
}
signed main() {
start;
cin >> T;
while (T--)
solve();
return 0;
}
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