2017 ACM/ICPC 沈阳 L题 Tree

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.

InputThe 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. 
OutputFor each test case, output the maximum size of E1 ∩ E1 ... ∩ 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
题解：考虑某条边，则只要两边的2个顶点都大于等于k,则连边时一定会经过这条边，ans++;

参看代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define clr(a,b,n) memset((a),(b),sizeof(int)*n)
 typedef long long ll;
 const int maxn = 2e5+;
 int n,k,ans,num[maxn];
 vector<int> vec[maxn];

 void dfs(int u,int fa)
 {
     num[u]=;
     for(int i=;i<vec[u].size();i++)
     {
         int v=vec[u][i];
         if(v==fa) continue;
         dfs(v,u);
         num[u]+=num[v];
         if(num[v]>=k&&n-num[v]>=k) ans++;
     }
 }

 int main()
 {
     int T,u,v;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&k);
         clr(num,,n+); ans=;
         for(int i=;i<=n;i++) vec[i].clear();
         for(int i=;i<n;i++)
         {
             scanf("%d%d",&u,&v);
             vec[u].push_back(v);
             vec[v].push_back(u);
         }
         dfs(,-);
         //for(int i=1;i<=n;++i) cout<<num[i]<<endl;
         printf("%d\n",ans);
     }
     return ;
 }