『算法-ACM 竞赛-CodeForces』1311E. Construct the Binary Tree （构造，就是个模拟)

ACM 思维题训练集合
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 在构造的树的深度最大(左偏或右偏树）和最小（满二叉树)之内就能构成，然后从左偏树开始不断的将低端的点向上移动，知道达到要求。

#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;
    } //1
    for (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;
}