【赛后补题】(HDU6228) Tree {2017-ACM/ICPC Shenyang Onsite}

这条题目当时卡了我们半天，于是成功打铁……今天回来一看，mmp，贪心思想怎么这么弱智。。。。。（怪不得场上那么多人A了

题意分析

这里是原题：

Tree

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

题意很简单，我们考虑一下做法。
要想交集尽可能大，颜色各自的分布应该尽可能“往顶层和底层分布”——这是贪心思想。然后具体怎么实现？很简单，对每一个点i，设它的子树的节点（包括它自身）有p个，那么只需要检查p≥k和n−p≥k即可。不需要查边，只需要查点，因为只要存在这样的点，那么一定存在这样的一条公共边。
实现查子树可以用dfs遍历一遍即可实现。

代码

#include <bits/stdc++.h>using namespace std;#define NQUICKIO#define NFILEstruct Edge{    int u,v;    Edge(int _u,int _v):u(_u),v(_v) {}};const int maxnode=200005;vector<Edge> edges;vector<int> G[maxnode];int s[maxnode];void addEdge(int u,int v){    edges.push_back(Edge(u,v));    G[u].push_back((int)edges.size()-1);    return;}int dfs(int f,int p){    //cout<<"now point:"<<p<<endl;    int nowsum=1;    for(int i=0;i!=(int)G[p].size();++i)        if(edges[G[p][i]].v!=f)            nowsum+=dfs(p,edges[G[p][i]].v);    return s[p]=nowsum;}int main(){#ifdef QUICKIO    ios::sync_with_stdio(false);    cin.tie(0); cout.tie(0);#endif#ifdef FILE    freopen("datain.txt","r",stdin);    freopen("dataout.txt","w",stdout);#endif    int T; cin>>T;    while(T--)    {        edges.clear();        int n,k; cin>>n>>k;        for(int i=1;i<=n;++i) G[i].clear();        memset(s,0,sizeof(s));        for(int i=1;i!=n;++i)        {            int tu,tv; cin>>tu>>tv;            addEdge(tu,tv);            addEdge(tv,tu);        }        dfs(-1,edges[0].u);        int ans=0;        /*for(int i=1;i<=n;++i)            cout<<s[i]<<" ";            cout<<endl;*/        for(int i=1;i<=n;++i)        {            if(s[i]>=k && n-s[i]>=k) ans++;        }        cout<<ans<<endl;    }    return 0;}