After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he sued Bessie to mitigate his losses.
Bessie, feeling vindictive, decided to sabotage Farmer John's network by cutting power to one of the barns (thereby disrupting all the connections involving that barn). When Bessie does this, it breaks the network into smaller pieces, each of which retains full connectivity within itself. In order to be as disruptive as possible, Bessie wants to make sure that each of these pieces connects together no more than half the barns on FJ.
Please help Bessie determine all of the barns that would be suitable to disconnect.

Input

* Line 1: A single integer, N. The barns are numbered 1..N.
* Lines 2..N: Each line contains two integers X and Y and represents a connection between barns X and Y.

Output

* Lines 1..?: Each line contains a single integer, the number (from 1..N) of a barn whose removal splits the network into pieces each having at most half the original number of barns. Output the barns in increasing numerical order. If there are no suitable barns, the output should be a single line containing the word "NONE".

Sample Input

Sample Output

3
8

Hint

INPUT DETAILS:
The set of connections in the input describes a "tree": it connects all the barns together and contains no cycles.

OUTPUT DETAILS:
If barn 3 or barn 8 is removed, then the remaining network will have one piece consisting of 5 barns and two pieces containing 2 barns. If any other barn is removed then at least one of the remaining pieces has size at least 6 (which is more than half of the original number of barns, 5).

题意：给了一棵节点数为n的树的对应关系，让判断去掉哪些节点能使子数的大小小于等于n/2，并将符合条件的节点从小到大输出，如果没有符合条件的节点，则输出 “NONE”。

思路：既然是让求哪些ji节点符合条件，可以用邻接表构建一个图，然后用dfs找出每一个节点下连的图的da'x大小些节点符合条件，将节点存入ans数组中。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 10010
using namespace std;
int n,s,e,first[N],book[N],ans[N];
struct node
{
    int x,y;
}que[2*N];
void add(int x,int y) //构建邻接表的函数
{
    que[e].x=x;
    que[e].y=first[y];
    first[y]=e++;
}
int dfs(int x)
{

    int num=0,sum=0;
    book[x]=1;
    int k=first[x],flag=0;
    while(k!=-1)
    {
        if(!book[que[k].x])
        {
            num=dfs(que[k].x);
            sum+=num;
            if(num>n/2)   //x节点下的图的大小超过了n/2，不符合条件
                flag=1;
        }
        k=que[k].y;
    }
    if(n-sum-1>n/2)    //另一半是否小于等于n/2
        flag=1;
    if(!flag)
        ans[s++]=x;
    return sum+1;     //加上自己
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(first,-1,sizeof(first));//初始化
        memset(book,0,sizeof(book));
        s=e=0;   //有s个节点符合条件，e用邻接表表示无向图的大小
        int a,b;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            add(a,b);       //构建邻接表
            add(b,a);       //无向图
        }
        a=dfs(1);        //从第一个点开始尝试
        if(s)      //有s个节点符合条件
        {
            sort(ans,ans+s);    //从小到大排序，输出
            for(int i=0;i<s;i++)
                printf("%d\n",ans[i]);
        }
        else    //没有符合条件的节点
            printf("NONE\n");
    }
}

Tree Cutting 题解