HDU 6228 - Tree - [DFS]
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6228
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
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
Source
2017ACM/ICPC亚洲区沈阳站-重现赛(感谢东北大学)
题意:
一颗无根树,有 $n$ 个节点,现在要给每个节点上 $k$ 种颜色中的一种,
然后对于第 $i$ 种颜色的所有节点,$E_i$ 是连接这些节点所需要的最少的边的集合;
要求最大的 $\left| {E_1 \cap E_2 \cap \cdots \cap E_{k - 1} \cap E_k } \right|$。
题解:
考虑对于任意一条边,把它的左侧节点 $u$ 看做统领一棵树,右侧节点 $v$ 也统领一棵树,两棵树的树根被当前这条边连接起来,
那么,不难想象,两侧树的节点数目均不小于 $k$ 是当前边作为“答案集”的一个元素的必要条件
(不过,我不知道怎么证明是充分条件,不过想来必然存在一种上色方案,使得上述必要条件成为充分条件),
那么,可以使用DFS遍历整棵树,对于遍历到的任意一条边,假设它连接到的子节点 $v$ 所统领的整棵子树的数目为 $cnt[v]$,
那么,只要满足 $cnt[v] \ge k$ 且 $n - cnt[v] \ge k$,当前这条边就算做“答案集”的一个元素,$ans++$;
最后,DFS结束,输出 $ans$ 即可。
AC代码:
#include<bits/stdc++.h>
using namespace std;
const int maxn=+; int n,k;
int ans; struct Edge{
int u,v;
Edge(int u=,int v=){this->u=u,this->v=v;}
};
vector<Edge> E;
vector<int> G[maxn];
void init(int l,int r)
{
E.clear();
for(int i=l;i<=r;i++) G[i].clear();
}
void addedge(int u,int v)
{
E.push_back(Edge(u,v));
G[u].push_back(E.size()-);
} int vis[maxn];
int dfs(int now)
{
vis[now]=;
int tot=;
for(int i=;i<G[now].size();i++)
{
Edge &e=E[G[now][i]]; int nxt=e.v;
if(!vis[nxt]) tot+=dfs(nxt);
}
if(n-tot>=k && tot>=k) ans++;
return tot;
} int main()
{
int T;
cin>>T;
while(T--)
{
scanf("%d%d",&n,&k);
init(,n);
for(int i=;i<n;i++)
{
int u,v;
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
} memset(vis,,sizeof(vis));
ans=;
dfs();
cout<<ans<<endl;
}
}
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