Tree Cutting(树的重心)

Tree Cutting

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

10
1 2
2 3
3 4
4 5
6 7
7 8
8 9
9 10
3 8

Sample Output

3
8

题目大意：

以每个节点为根，统计出每个节点的最大子树节点个数，找出小于等于n/2的所有节点。

题目解析：

运用找树的重心的方法，统计出每个节点的最大子树的节点个数，判断一下就行。

代码如下：

#include 
#include 
using namespace std;

const int maxn = 10000 + 5;

struct Edge{
    int next,to;
}G[maxn << 1];

int n,head[maxn],cnt;
int f[maxn],dp[maxn];

void Insert(int u,int v){
    G[++cnt].next = head[u];
    G[cnt].to = v;
    head[u] = cnt;
}

void dfs(int u,int fa){
    f[u] = 1;
    int maxd = 0;
    for(int e = head[u];e;e = G[e].next){
        int v = G[e].to;
        if(v == fa)continue;
        dfs(v,u);
        f[u] += f[v];
        maxd = max(maxd,f[v]);
    }
    dp[u] = max(maxd,n - f[u]);
}

void init(){
    for(int i = 0;i < maxn; i++){
        dp[i] = 0;
        head[i] = 0;
        f[i] = 0;
    }
    cnt = 0;
}

int main(){
    int u,v;
    while(~scanf("%d",&n) && n){
        init();
        for(int i = 0;i < n-1; i++){
            scanf("%d%d",&u,&v);
            Insert(u,v);
            Insert(v,u);
        }
        dfs(1,-1);
        int ans = 0;
        for(int i = 1;i <= n; i++){
            if(dp[i] <= n/2){
                printf("%d\n",i);
            }
        }
    }
}