hdu6228Tree

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

1 0 1

 

中文题意：给你一个数T,代表case的个数；

然后每个case,给两个数node(点的个数)，k(颜色的个数)，接下来node-1行，每行两个数，代表这两个数之间存在edge;

然后保证每次case都是一棵树，现在用这k种颜色为树的节点进行染色，Ei代表第i种颜色的所有节点之间相连的所有边的集合；

E[1],E[2],E[3]...E[node]中公共边数为最大数量；

 

感想：第一次做这道题时，有点懵，想了一阵之后，脑海中有一点思路，既然是要求边，那就从边进行研究，可是脑海中的那一点灵光总是转不住，然后就在网上搜了题解，结果题解跟自己的那点灵光很像，但还是有一点地方有点疑惑；最后的AC代码还是在询问队友之后了解到了一些知识点采写出，总的来说我还是太菜了！！！需要努力！

 

思路：对于某一个结点来说，如果它的左边（算上自身）能有k个结点的话，而它的右边也能有k个节点的话，那么这个结点右边的这条边就会在所有的边集中。

 

AC代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=200010;
vector<int >ve[maxn];
int plug[maxn];
int sum[maxn];
int number(int x){ //对于这里额使用的是一个数组来标记这个点是否访问过，其实也可以直接将上一个访问的点放进来进行判断就好，因为这个图也是一棵树
for(int i=0;i<ve[x].size();i++){
if(plug[ve[x][i]]) continue;
plug[ve[x][i]]=1;
sum[x]+=number(ve[x][i]);
}
return sum[x];
}
int main(){
int T,node,k,x,y;
cin>>T;
while(T--){
scanf("%d%d",&node,&k);
for(int i=0;i<=maxn+5;i++) ve[i].clear();
memset(plug,0,sizeof(plug));
for(int i=0;i<=node;i++) sum[i]=1;
for(int i=0;i<node-1;i++){
scanf("%d%d",&x,&y);
ve[x].push_back(y);
ve[y].push_back(x);
}
plug[1]=1;
number(1);
/*for(int i=1;i<node;i++) printf("%d ",sum[i]);
printf("%d ",sum[node]);*/
int ans=0;
for(int i=1;i<=node;i++){
if(sum[i]>=k&&node-sum[i]>=k) ans++;
}
printf("%d ",ans);
}
}