hdu 6228 Tree

TreeTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 539    Accepted Submission(s): 343

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

34 21 22 33 44 21 21 31 46 31 22 33 43 56 2

 

Sample Output

101

 

Source

2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

题意：

让你给一个n个节点的树形图涂色，有k中颜色，Ei表示最小的边集，使得所有涂颜色i的点连通。让你选定一种涂色方案，使得res=E1∩E2∩。。。∩Ek最大

解析：

将问题局部化，考虑一条边连接两个点，假若A,B有边，且A边的子树种节点数为w+1（包括A）,那么B边的子树的大小就是n-w-1(包括B),若两边子树的大小都大于k的话，那么这一定是一条所有颜色的公共边，即答案+1.这样用DFS就可以预处理出子树大小，然后在遍历边就可以得出答案。

#include<cstdio>#include<vector>#include<cstring>using namespace std;const int MAXN = 200000+10;int n,k;vector<int> qq[MAXN];int sont[MAXN];int visit[MAXN];int edge[MAXN][2];int  dfs(int x,int depth){    visit[x]=depth;    int num=0;    for(int i=0;i<qq[x].size();i++)    {        int u=qq[x][i];        if(visit[u]==0)        {            num+=dfs(u,depth+1);        }    }    sont[x]=num;    return num+1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&k);        for(int i=1;i<=n;i++)        {            qq[i].clear();        }        for(int i=1;i<n;i++)        {            int u,v;            scanf("%d%d",&u,&v);            edge[i][0]=u;            edge[i][1]=v;            qq[u].push_back(v);            qq[v].push_back(u);        }        memset(visit,0,sizeof(visit));        dfs(1,1);        int ans=0;        for(int i=1;i<n;i++)        {            int u=edge[i][0];            int v=edge[i][1];            if(visit[u]>visit[v])            {                if(sont[u]+1>=k&&sont[u]+1<=n-k)                {                    ans++;                }            }            else            {                if(sont[v]+1>=k&&sont[v]+1<=n-k)                {                    ans++;                }            }        }        printf("%d\n",ans);    }    return 0;}