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题解：考虑某条边，则只要两边的2个顶点都大于等于k,则连边时一定会经过这条边，ans++; 参看代码：
#include<bits/stdc++.h>
using namespace std;
#define clr(a,b,n) memset((a),(b),sizeof(int)*n)
typedef long long ll;
const int maxn = 2e5+10;
int n,k,ans,num[maxn];
vector<int> vec[maxn];
void dfs(int u,int fa)
{
num[u]=1;
for(int i=0;i<vec[u].size();i++)
{
int v=vec[u][i];
if(v==fa) continue;
dfs(v,u);
num[u]+=num[v];
if(num[v]>=k&&n-num[v]>=k) ans++;
}
}
int main()
{
int T,u,v;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
clr(num,0,n+1); ans=0;
for(int i=1;i<=n;i++) vec[i].clear();
for(int i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
vec[u].push_back(v);
vec[v].push_back(u);
}
dfs(1,-1);
//for(int i=1;i<=n;++i) cout<<num[i]<<endl;
printf("%d\n",ans);
}
return 0;
}
原文地址：https://www.cnblogs.com/songorz/p/9784973.html