HDU 6228/2017ACM/ICPC 沈阳 Tree 【DFS】

Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit:262144/262144 K (Java/Others)
Total Submission(s): 19    Accepted Submission(s): 10

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 1to n. If you cannot understand the concept of a tree here, please omit thisproblem.
Now we decide to colour its nodes with k distinct colours, labelled from 1 tok. Then for each colour i = 1, 2, · · · , k, define Ei as the minimum subset ofedges connecting all nodes coloured by i. If there is no node of the treecoloured by a specified colour i, Ei will be empty.
Try to decide a colour scheme to maximize the size of E1 ∩ E2 · · · ∩ Ek, andoutput its size.

 

 

Input

The first line of input contains an integer T (1 ≤ T ≤ 1000), indicating the totalnumber of test cases.
For each case, the first line contains two positive integers n which is thesize of the tree and k (k ≤ 500) which is the number of colours. Each of thefollowing n - 1 lines contains two integers x and y describing an edge betweenthem. 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

【题意】

给出一棵有n个节点的树，现在你可以用k种颜色对节点染色，每种颜色对应一个集合，表示将树上所有这种颜色的点连起来经过的最小边。现在需要求所有集合取交集后的大小。

【思路】

假设我们取定1为根节点。

显然要是结果最大，相同颜色应该要尽可能分布在树的顶部和底部。

虽然我们需要考虑的是边，但是我们可以转化为对每个节点去考虑。令在这个节点的两侧每种颜色均有分布，那么一定会有一条公共边。

先用DFS求出每个点的子孙节点个数(包括自身)，假设为x，那么它的上面点的个数为n-x,只要x>=k&&(n-x)>=k即能满足上面的条件。

我们只要枚举每个点，满足条件点的个数即为答案。

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef long long ll;
const int maxn = 200005;
const ll mod = 10;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;

int n,k,ans;
int num[maxn];
vectorvec[maxn];

void dfs(int u,int pre)
{
    num[u]=1;
    for(int i=0;i=k&&n-num[v]>=k) ans++;
    }
}

int main()
{
    int u,v;
    rush()
    {
        mst(num,0);
        scanf("%d%d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            vec[i].clear();
        }
        for(int i=1;i