原题
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
解析
题目英文比较易懂就不说大意了。
思路是考虑每一条边，如果此边两侧的点数都大于k，则这个边就可以存在于并集中。遍历所有边计数即可。
AC代码
#include<bits/stdc++.h>
using namespace std;
int const maxn=200000+5;
int n,k;
#define register int reg
vector<int>mp[maxn];
int sum;
int dfs(int rt,int pre){
int sz=mp[rt].size();
int cnt=0;
for(int i=0;i<sz;i++){
int c=mp[rt][i];
if(c!=pre){
c=dfs(c,rt);
cnt+=c;
if(c>=k && n-c>=k){
sum++;
}
}
}
return cnt+1;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("r.txt","r",stdin);
#endif
int T;
scanf("%d",&T);
int i;
while(T--){
sum=0;
scanf("%d%d",&n,&k);
for(i=1;i<=n;i++){
mp[i].clear();
}
int a,b;
for(i=1;i<n;i++){
scanf("%d%d",&a,&b);
mp[b].push_back(a);
mp[a].push_back(b);
}
dfs(1,-1);
printf("%d\n",sum);
}//end of while
}