Justified Jungle

Problem J: Justified Jungle

Time limit: 6 s Memory l

imit: 512 MiB

As you probably know, a tree is a graph consisting of n nodes and n−1 undirected edges in which any two nodes are connected by exactly one path. A forest is a graph consisting of one or more trees. In other words, a graph is a forest if every connected component is a tree. A forest is justified if all connected components have the same number of nodes. Given a tree G consisting of n nodes, find all positive integers k such that a justified forest can be obtained by erasing exactly k edges from G. Note that erasing an edge never erases any nodes. In particular when we erase all n−1 edges from G, we obtain a justified forest consisting of n one-node components.
Input The first line contains an integer n (2≤ n ≤1000000) — the number of nodes in G. The k-th of the following n−1 lines contains two different integers ak and bk (1≤ ak,bk ≤n) — the endpoints of the k-th edge.
Output
The first line should contain all wanted integers k, in increasing order.
Example
input
8 1 2 2 3 1 4 4 5 6 7 8 3 7 3
output
1 3 7
Figures depict justified forests obtained by erasing 1, 3 and 7 edges from the tree in the example input.

 // 题目大意：删去k条边，树变为相等个点的连通分量，求所有正整数k。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 #define maxn 1000006
 int head[maxn],cnt,siz[maxn],v[maxn],n;
 struct edge{
     int to,nxt;
 }e[maxn<<];
 void add_edge(int u,int v){
     e[cnt].to=v;
     e[cnt].nxt=head[u];
     head[u]=cnt++;
 }
 void dfs(int u,int fa){
     siz[u]=;
     for(int i=head[u];i!=-;i=e[i].nxt){
         int v=e[i].to;
         if(v==fa) continue;
         dfs(v,u);
         siz[u]+=siz[v];
     }
     v[siz[u]]++;
 }
 bool check(int x){
     x++;
     if(n%x) return ;
     int w=n/x,sum=;
     for(int i=w;i<=n;i+=w) sum+=v[i];
     return sum==x;
 }
 int main(){
     memset(head,-,sizeof head);
     scanf("%d",&n);
     for(int i=;i<n-;i++){
         int a,b;
         scanf("%d%d",&a,&b);
         add_edge(a,b);
         add_edge(b,a);
     }
     dfs(,-);
     for(int i=;i<=n;i++){
         if(check(i)) printf("%d ",i);
     }
     return ;
 }