can 可能 樹的重心 test sam example center trees esp

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Balancing Act

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14582		Accepted: 6192

Description

Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T.
For example, consider the tree:

Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two.

For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.

Input

The first line of input contains a single integer t (1 <= t <= 20), the number of test cases. The first line of each test case contains an integer N (1 <= N <= 20,000), the number of congruence. The next N-1 lines each contains two space-separated node numbers that are the endpoints of an edge in the tree. No edge will be listed twice, and all edges will be listed.

Output

For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.

Sample Input

1
7
2 6
1 2
1 4
4 5
3 7
3 1

Sample Output

1 2
題意大概就是給你一棵樹，讓你找樹的重心。*樹的重心：在一顆樹中找到一個節點，這個節點的最大子樹相對於其他節點的最大子樹最小，該節點就是樹的重心。刪去重心，生成的一堆樹盡可能平衡。
思路大概就是dfs遍歷每一個點，然後找到該點的最大子樹，不斷更新擁有最小的最大子樹的節點。
代碼如下：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define maxn 20010
 5 #define mmst(x,y) memset(x,y,sizeof(x))
 6 using namespace std;
 7 int son[maxn],size,ans,T,n,ecnt,head[maxn];
 8 bool vis[maxn];
 9 struct edge
10 {
11     int u,v,next;
12 }E[maxn*2];
13 void add(int u,int v)
14 {
15     E[ecnt++].u=u;
16     E[ecnt].v=v;
17     E[ecnt].next=head[u];
18     head[u]=ecnt;
19 }
20 void dfs(int u)
21 {
22     vis[u]=1;
23     son[u]=0;
24     int tmp=0;
25     for(int i=head[u];i;i=E[i].next)
26     {
27         int v=E[i].v;
28         if(vis[v]) continue;
29         dfs(v);
30         son[u]+=son[v]+1;
31         tmp=max(tmp,son[v]+1);
32     }
33     tmp=max(tmp,n-son[u]-1);    
34     if(tmp<size||tmp==size&&u<ans)
35     {
36         size=tmp;
37         ans=u;
38     }
39 }
40 void clr()
41 {
42     mmst(son,0);
43     mmst(head,0);
44     mmst(vis,0);
45     ecnt=0;
46     ans=size=0x3f3f3f3f;
47 }
48 int main()
49 {
50     scanf("%d",&T);
51     while(T--)
52     {
53         clr();
54         scanf("%d",&n);
55         for(int i=1;i<n;++i)
56         {
57             int a,b;
58             scanf("%d%d",&a,&b);
59             add(a,b);
60             add(b,a);
61         }
62          dfs(1);
63         printf("%d %d\n",ans,size);
64     }
65 }

poj1655 Balancing Act