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
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 分析: 问题可以转化为求有多少条边左右两边都有k个以上的点。因为这样的边一定是符合要求的。题目保证这是一个树,树是无向无环图,所以我们可以从一个点开始DFS搜索,进而找到每一个点左右两边的点数 代码如下:
#include<cstdio> #include<iostream> #include<algorithm> #include<vector> #include<cstring> using namespace std; const int MAXN=2e5+10; int vis[MAXN]; int num[MAXN]; int cnt; vector<int>V[MAXN]; struct node { int u; int v; }poi[MAXN]; int t,n,k,a,b,ans; void dfs(int x) { int u,v; num[x]=1; for(int i=0;i<V[x].size();i++) { u=x; v=V[x][i]; /* if(v==pre) continue;*/ if(!vis[v]) { vis[v]=1; dfs(v); num[u]+=num[v]; } } if(num[u]>=k&&n-num[u]>=k) ans++; } int main() { scanf("%d",&t); while(t--) { memset(vis,0,sizeof(vis)); ans=0; scanf("%d%d",&n,&k); for(int i=0;i<=n;i++)V[i].clear(); int h=n-1; for(int i=0;i<n-1;i++){ scanf("%d%d",&poi[i].u,&poi[i].v); V[poi[i].u].push_back(poi[i].v); V[poi[i].v].push_back(poi[i].u); } vis[1]=1; dfs(1); printf("%d\n",ans); } return 0; }