ACM思維題訓練集合
You are given two integers n and d. You need to construct a rooted binary tree consisting of n vertices with a root at the vertex 1 and the sum of depths of all vertices equals to d.

A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. A parent of a vertex v is the last different from v vertex on the path from the root to the vertex v. The depth of the vertex v is the length of the path from the root to the vertex v. Children of vertex v are all vertices for which v is the parent. The binary tree is such a tree that no vertex has more than 2 children.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤1000) — the number of test cases.

The only line of each test case contains two integers n and d (2≤n,d≤5000) — the number of vertices in the tree and the required sum of depths of all vertices.

It is guaranteed that the sum of n and the sum of d both does not exceed 5000 (∑n≤5000,∑d≤5000).

Output
For each test case, print the answer.

If it is impossible to construct such a tree, print “NO” (without quotes) in the first line. Otherwise, print “{YES}” in the first line. Then print n?1 integers p2,p3,…,pn in the second line, where pi is the parent of the vertex i. Note that the sequence of parents you print should describe some binary tree.

Example
inputCopy
3
5 7
10 19
10 18
outputCopy
YES
1 2 1 3
YES
1 2 3 3 9 9 2 1 6
NO
Note
Pictures corresponding to the first and the second test cases of the example:

丫的，改了一天。
如果b在構造的樹的深度最大(左偏或右偏樹）和最小（滿二叉樹)之內就能構成，然后從左偏樹開始不斷的將低端的點向上移動，知道達到要求。

#include <bits/stdc++.h> using namespace std; int f[210]; inline void solve() {memset(f, 0, sizeof(f));int n, d, maxd = 0;scanf("%d %d", &n, &d);--n;if (d > n * (n + 1) / 2){printf("NO\n");return;} //1for (int i = 1;; ++i){maxd = i;if (n > (1 << i)){d -= i * (1 << i);f[i] = 1 << i;n -= 1 << i;}else{d -= i * n;f[i] = n;n -= n;break;}}if (d < 0){printf("NO\n");return;}while (1){if (d == 0)break;int p;for (p = maxd; p >= 1; --p)if (f[p] > 1)break;--d;--f[p];++f[p + 1];if (p + 1 > maxd)maxd = p + 1;}printf("YES\n");int p = 1, np = 1, cnt;for (int i = 1; i <= maxd; ++i){int t = p;cnt = 0;for (int j = 1; j <= f[i]; ++j){++p;++cnt;if (cnt >= 3){++np;cnt = 1;} printf("%d ", np);}np = t + 1;}printf("\n"); } int main() {int t;scanf("%d", &t);for (int i = 1; i <= t; ++i)solve();return 0; } 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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