2017 ICPC/ACM 沈阳区域赛HDU6228

Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 373    Accepted Submission(s): 254

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 ∩ 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

1

题意 给树节点染色

直接dfs搜一遍就好了

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <ctime>
 #include <vector>
 using namespace std;
 const int maxn= 1e5+;
 const int maxm= 1e6+;
 const int inf = 0x3f3f3f3f;
 typedef long long ll;
 vector<int> G[maxn];
 int n,k,ans;
 int siz[maxn];
 void dfs(int x,int fa)
 {
     siz[x]=;
     for(int i=;i<G[x].size();i++)
     {
         int w=G[x][i];
         if(w!=fa)
         {
             dfs(w,x);
             siz[x]+=siz[w];
         }
     }
     if(siz[x]>=k&&n-siz[x]>=k)
         ans++;
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&k);
         int u,v;
         for(int i=;i<=n;i++)
             G[i].clear();
         for(int i=;i<n;i++)
         {
             scanf("%d%d",&u,&v);
             G[u].push_back(v);
             G[v].push_back(u);
         }
         ans=;
         dfs(,);
         printf("%d\n",ans);
     }
 }