E. Construct the Binary Tree

题目大意：

要求构造一个n个节点的二叉树（每个节点拥有不超过2个孩子），节点1为根，要使所有节点到根的距离之和为d。要求先判断可不可以构造，如果可以输出“YES”，下一行输出2到n号节点的父亲节点，否则输出“NO”。有多组询问。

 

思路：

首先不难想到深度和最小的那种就是满二叉树的形式。然后我们再考虑移动一个节点到下一层去。 就是这么简单

#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <math.h>
#include <cstdio>
#include <iomanip>
#include <time.h>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>

#define LL long long
#define INF 0x3f3f3f3f
#define ls nod<<1
#define rs (nod<<1)+1

const double eps = 1e-10;
const int maxn = 5e3 + 10;;
const LL mod = 1e9 + 7;

int sgn(double a){return a < -eps ? -1 : a < eps ? 0 : 1;}
using namespace std;


int fa[maxn],dep[maxn];
vector<int> vec[maxn];
int main() {
    int T;
    cin >> T;
    while (T--) {
        int n,m;
        cin >> n >> m;
        for (int i = 1;i <= n;i++)
            dep[i] = 0,fa[i] = 0,vec[i].clear();
        vec[0].push_back(1);
        int maxx = 0;
        for (int i = 2;i <= n;i++) {
            dep[i] = dep[i>>1] + 1;
            vec[dep[i]].push_back(i);
            maxx = max(maxx,dep[i]);
            m -= dep[i];
        }
        if (m < 0) {
            cout << "NO" << endl;
            continue;
        }
        bool flag = true;
        while (m) {
            int change = maxx;
            while (change >= 0 && vec[change].size() == 1)
                change--;
            if (change <= 0) {
                cout << "NO" << endl;
                flag = false;
                break;
            }
            int nod = vec[change][vec[change].size()-1];
            vec[change].pop_back();
            vec[change+1].push_back(nod);
            maxx = max(maxx,change+1);
            m--;
        }
        if (flag) {
            cout << "YES" << endl;
            for (int i = 1;i <= n;i++) {
                for (int j = 0;j < vec[i].size();j++)
                    fa[vec[i][j]] = vec[i-1][j>>1];
            }
            for (int i = 2;i <= n;i++) {
                cout << fa[i] << " ";
            }
            cout << endl;
        }
    }
    return 0;
}