HDU6228-Tree

Tree

                                                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

                                                                                                Total Submission(s): 413    Accepted Submission(s): 281

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亞洲區沈陽站-重制賽（感謝東北大學）  

Recommend jiangzijing2015  

題意：一棵樹有n個點，有k種顔色，對這n個點進行染色，對于每種顔色，把染成這種顔色的點連接配接起來的邊拿出來形成一個集合，問這k個集合交集的最大值為多少

解題思路：因為交集要盡量大，是以必然k中顔色要染在盡量小的且大小大于等于k的子樹中。先找到樹的重心，然後k個k個染即可

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int s[200009],nt[4000009],e[400009],cnt;
int n,m,u,v,tot;
int sum[200009], mx[200009], vis[200009];

int dfs(int k, int fa, int p)
{
    int ans = 0;
    sum[k] = 1;  mx[k] = 0;
    for (int i = s[k]; ~i; i = nt[i])
    {
        if (e[i] == fa || vis[e[i]]) continue;
        int temp = dfs(e[i], k, p);
        sum[k] += sum[e[i]];
        if(sum[e[i]]>=m) tot++;
        mx[k] = max(mx[k], sum[e[i]]);
        if (mx[temp] < mx[ans]) ans = temp;
    }
    mx[k] = max(mx[k], p - sum[k]);
    return mx[k] < mx[ans] ? k : ans;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(s,-1,sizeof s);
        mx[cnt=0]=INF;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            nt[cnt]=s[u],s[u]=cnt,e[cnt++]=v;
            nt[cnt]=s[v],s[v]=cnt,e[cnt++]=u;
        }
        int y=dfs(1,1,n);
        tot=0;
        dfs(y,y,n);
        printf("%d\n",tot);
    }
    return 0;
}