Balancing Act

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 1655
64-bit integer IO format: %lld      Java class name: Main
 
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: 
POJ 1655 Balancing Act_#include

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

Source

 
解题：求树的重心
 
POJ 1655 Balancing Act_ios_02POJ 1655 Balancing Act_ios_03
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int maxn = 20010;
 6 struct arc {
 7     int to,next;
 8     arc(int x = 0,int y = -1) {
 9         to = x;
10         next = y;
11     }
12 } e[maxn<<1];
13 int head[maxn],siz[maxn],tot,n;
14 void add(int u,int v) {
15     e[tot] = arc(v,head[u]);
16     head[u] = tot++;
17 }
18 int ret,idx;
19 void dfs(int u,int fa) {
20     siz[u] = 1;
21     int tmp = 0;
22     for(int i = head[u]; ~i; i = e[i].next) {
23         if(e[i].to == fa) continue;
24         dfs(e[i].to,u);
25         tmp = max(siz[e[i].to],tmp);
26         siz[u] += siz[e[i].to];
27     }
28     tmp = max(tmp,n - siz[u]);
29     if(tmp < ret || tmp == ret && u < idx) {
30         ret = tmp;
31         idx = u;
32     }
33 }
34 int main() {
35     int kase,u,v;
36     scanf("%d",&kase);
37     while(kase--) {
38         scanf("%d",&n);
39         memset(head,-1,sizeof head);
40         idx = ret = n<<1;
41         tot = 0;
42         for(int i = 1; i < n; ++i) {
43             scanf("%d%d",&u,&v);
44             add(u,v);
45             add(v,u);
46         }
47         dfs(1,-1);
48         printf("%d %d\n",idx,ret);
49     }
50     return 0;
51 }
View Code

 

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