Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1693 Accepted Submission(s): 978

Problem Description

Consider a un-rooted tree T which is not the biological significance of tree or plant, but a tree as an undirected graph in graph theory with n nodes, labelled from 1 to n. If you cannot understand the concept of a tree here, please omit this problem.
Now we decide to colour its nodes with k distinct colours, labelled from 1 to k. Then for each colour i = 1, 2, · · · , k, define Ei as the minimum subset of edges connecting all nodes coloured by i. If there is no node of the tree coloured by a specified colour i, Ei will be empty.
Try to decide a colour scheme to maximize the size of E1 ∩ E2 · · · ∩ Ek, and output its size.

Input

The first line of input contains an integer T (1 ≤ T ≤ 1000), indicating the total number of test cases.
For each case, the first line contains two positive integers n which is the size of the tree and k (k ≤ 500) which is the number of colours. Each of the following n - 1 lines contains two integers x and y describing an edge between them. We are sure that the given graph is a tree.
The summation of n in input is smaller than or equal to 200000.

Output

For each test case, output the maximum size of E1 ∩ E2 ... ∩ Ek.

Sample Input

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

Sample Output

1 0 1

给你n个节点，k个颜色，要你用k个颜色去涂这n个节点。E_i表示将所有颜色为i的结点连起来的最小边数。E1 ∩ E2 ... ∩ Ek表示E1 E2...Ek的重合边数，输出最大的E1 ∩ E2 ... ∩ Ek。

求出每个节点的子树大小（包括自己），如果子树大小大于等于k并且n-子树大小也大于等于k，ans+1。

#include<iostream>
#include<vector>
using namespace std;
#define maxn 300000
int n,k,cnt,ans,size[maxn],head[maxn];
struct edge{int to,next;
}e[maxn];
vector<int>ve[maxn];
void add(int u,int v)
{e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;
}
void dfs(int u,int f)
{for(int i=0;i<ve[u].size();i++){int x=ve[u][i];if(x==f)continue;dfs(x,u);size[u]+=size[x];}if(size[u]>=k&&n-size[u]>=k)ans++; 
}
int main()
{int t;cin>>t;while(t--){cin>>n>>k;int u,v;for(int i=1;i<=n;i++){ve[i].clear();size[i]=1;}for(int i=1;i<n;i++){cin>>u>>v;add(u,v);ve[u].push_back(v);ve[v].push_back(u);}ans=0;dfs(1,0);cout<<ans<<endl;}return 0;
}