2017 ICPC/ACM 沈阳区域赛HDU6228

Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 373    Accepted Submission(s): 254

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

题意 给树节点染色

直接dfs搜一遍就好了

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <queue>
#include <ctime>
#include <vector>
using namespace std;
const int maxn= 1e5+5;
const int maxm= 1e6+5;
const int inf = 0x3f3f3f3f;
typedef long long ll;
vector<int> G[maxn];
int n,k,ans;
int siz[maxn];
void dfs(int x,int fa)
{
    siz[x]=1;
    for(int i=0;i<G[x].size();i++)
    {
        int w=G[x][i];
        if(w!=fa)
        {
            dfs(w,x);
            siz[x]+=siz[w];
        }
    }
    if(siz[x]>=k&&n-siz[x]>=k)
        ans++;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        int u,v;
        for(int i=1;i<=n;i++)
            G[i].clear();
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        ans=0;
        dfs(1,0);
        printf("%d
",ans);
    }
}