poj1655 Balancing Act

Posted 2020-10-08

---恢复内容开始---

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
题意大概就是给你一棵树，让你找树的重心。*树的重心：在一颗树中找到一个节点，这个节点的最大子树相对于其他节点的最大子树最小，该节点就是树的重心。删去重心，生成的一堆树尽可能平衡。
思路大概就是dfs遍历每一个点，然后找到该点的最大子树，不断更新拥有最小的最大子树的节点。
代码如下：

 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 }