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

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 ;

 }

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2017 ACM/ICPC 沈阳 L题 Tree

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