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2017沈阳站 Tree

发布时间:2024/7/5 编程问答 112 豆豆
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6228

Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1693    Accepted Submission(s): 978


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 ∩ E2 ... ∩ 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 给你n个节点,k个颜色,要你用k个颜色去涂这n个节点。Ei表示将所有颜色为i的结点连起来的最小边数。E1 ∩ E2 ... ∩ Ek表示E1 E2...Ek的重合边数,输出最大的E1 ∩ E2 ... ∩ Ek。 求出每个节点的子树大小(包括自己),如果子树大小大于等于k并且n-子树大小也大于等于k,ans+1。 #include<iostream> #include<vector> using namespace std; #define maxn 300000 int n,k,cnt,ans,size[maxn],head[maxn]; struct edge{int to,next; }e[maxn]; vector<int>ve[maxn]; void add(int u,int v) {e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt; } void dfs(int u,int f) {for(int i=0;i<ve[u].size();i++){int x=ve[u][i];if(x==f)continue;dfs(x,u);size[u]+=size[x];}if(size[u]>=k&&n-size[u]>=k)ans++; } int main() {int t;cin>>t;while(t--){cin>>n>>k;int u,v;for(int i=1;i<=n;i++){ve[i].clear();size[i]=1;}for(int i=1;i<n;i++){cin>>u>>v;add(u,v);ve[u].push_back(v);ve[v].push_back(u);}ans=0;dfs(1,0);cout<<ans<<endl;}return 0; }

 

转载于:https://www.cnblogs.com/chen99/p/10706615.html

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