标签：out acm minimum math.h back roo for nec tin
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
题意 给树节点染色
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 沈阳区域赛HDU6228
标签：out acm minimum math.h back roo for nec tin
原文地址：http://www.cnblogs.com/stranger-/p/7841226.html
