Codeforces-980E：The Number Games（倍增）

E. The Number Games

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The nation of Panel holds an annual show called The Number Games, where each district in the nation will be represented by one contestant.

The nation has $n$ districts numbered from $1$ to $n$, each district has exactly one path connecting it to every other district. The number of fans of a contestant from district $i$ is equal to $2^i$.

This year, the president decided to reduce the costs. He wants to remove $k$ contestants from the games. However, the districts of the removed contestants will be furious and will not allow anyone to cross through their districts.

The president wants to ensure that all remaining contestants are from districts that can be reached from one another. He also wishes to maximize the total number of fans of the participating contestants.

Which contestants should the president remove?

Input

The first line of input contains two integers $n$ and $k$ ($1\le k<n\le {10}^6$) — the number of districts in Panel, and the number of contestants the president wishes to remove, respectively.

The next $n-1$ lines each contains two integers $a$ and $b$ ($1\le a,b\le n$, $a\ne b$), that describe a road that connects two different districts $a$and $b$ in the nation. It is guaranteed that there is exactly one path between every two districts.

Output

Print $k$ space-separated integers: the numbers of the districts of which the contestants should be removed, in increasing order of district number.

Examples

input

Copy

6 3
2 1
2 6
4 2
5 6
2 3

output

Copy

1 3 4

input

Copy

8 4
2 6
2 7
7 8
1 2
3 1
2 4
7 5

output

Copy

1 3 4 5

Note

In the first sample, the maximum possible total number of fans is $2^2+2^5+2^6=100$. We can achieve it by removing the contestants of the districts 1, 3, and 4.

思路：删除点比较困难，可以保留点。很明显要尽量保留较大的点，因为2^n大于其余所有节点的值的和。那么以n为根节点建一棵树。

然后贪心保留n-1这个点，判断n-1这个点往根节点的第一个保留下来的节点之间路径的节点数是否小于或等于剩下的可以保留的节点数，若小于或等于，那么就把这条链上的点都保留下来，否则继续处理n-2这个点，以此类推。如果要以某个点为起点往上查找第一个被保留下来的点，就要利用树上的倍增了。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e6+10;
const int MOD=1e9+7;
const double PI=acos(-1.0);
typedef long long ll;
vector<int>e[MAX];
int nex[MAX][22],v[MAX];//v[i]判断i是否被保留下来
void dfs(int k,int fa)  //预处理nex数组
{
    for(int i=0;i<e[k].size();i++)
    {
        if(e[k][i]==fa)continue;
        nex[e[k][i]][0]=k;
        for(int j=1;j<=20;j++)nex[e[k][i]][j]=nex[nex[e[k][i]][j-1]][j-1];
        dfs(e[k][i],k);
    }
}
int check(int x,int tot)
{
    for(int i=20;i>=0;i--)
    {
        if(v[nex[x][i]]==0)
        {
            tot-=(1<<i);
            x=nex[x][i];
        }
    }
    return tot>=1;
}
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<n;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    for(int i=0;i<=20;i++)nex[n][i]=n;
    dfs(n,n);
    v[n]=1;
    k=n-k-1;
    for(int i=n-1;i>=1;i--)//从大到小枚举要保留的点
    {
        if(check(i,k))
        {
            int next=i;
            while(v[next]==0)//把当前点到第一个被保留的点之间的点都保留下来
            {
                v[next]=1;
                next=nex[next][0];
                k--;
            }
        }
    }
    for(int i=1;i<=n;i++)if(v[i]==0)printf("%d ",i);
    return 0;
}