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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 0 1

題意 給樹節點染色

直接dfs搜一遍就好了

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5
 
 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <ctime>
11 #include <vector>
12 using namespace std;
13 const int maxn= 1e5+5;
14 const int maxm= 1e6+5;
15 const int inf = 0x3f3f3f3f;
16 typedef long long ll;
17 vector<int> G[maxn];
18 int n,k,ans;
19 int siz[maxn];
20 void dfs(int x,int fa)
21 {
22     siz[x]=1;
23     for(int i=0;i<G[x].size();i++)
24     {
25         int w=G[x][i];
26         if(w!=fa)
27         {
28             dfs(w,x);
29             siz[x]+=siz[w];
30         }
31     }
32     if(siz[x]>=k&&n-siz[x]>=k)
33         ans++;
34 }
35 int main()
36 {
37     int t;
38     scanf("%d",&t);
39     while(t--)
40     {
41         scanf("%d%d",&n,&k);
42         int u,v;
43         for(int i=1;i<=n;i++)
44             G[i].clear();
45         for(int i=1;i<n;i++)
46         {
47             scanf("%d%d",&u,&v);
48             G[u].push_back(v);
49             G[v].push_back(u);
50         }
51         ans=0;
52         dfs(1,0);
53         printf("%d\n",ans);
54     }
55 }

2017 ICPC/ACM 沈陽區域賽HDU6228