POJ 1655 Balancing Art

 

Balancing Act

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4573		Accepted: 1800

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

Source

POJ Monthly--2004.05.15 IOI 2003 sample task

 

/* DFS树形DP搞定，minBal = min(minBal, max(n - num(i), num(k) for all i's son k)) for all nodes i 一开始采取的方案是：边输入边统计节点子孙数量，最后进行一遍线性扫描即可， 但是一直WA，后来分析了一下发现问题出在输入数据上的，输入数据并不一定是按照树形结构来的， 如果边输入边统计节点子孙数量那么很明显这就默认输入是按照树形结构来的，所以出错。后来改成 DFS就没有这个问题了 何为非树形结构输入：举个例子，对于树1->2->3，树形结构输入为1 2， 2 3；非树形结构输入为 1 2， 3 2这个特点决定了你不可能边输入边统计每个节点的子孙信息的 */ #include <iostream> #include <vector> #define maxv(a, b) ((a) >= (b) ? (a) : (b)) #define minv(a, b) ((a) <= (b) ? (a) : (b)) #define MAX_N 20010 using namespace std; struct node { vector<int> sons; }nodes[MAX_N + 5]; bool v[MAX_N + 5]; int n, minBal, minId; void init() { for(int i = 1; i <= n; i++) { v[i] = false; nodes[i].sons.clear(); } } int dfs(int curNode) { int curMax = INT_MIN, toId, total = 0; vector<int>::iterator iter = nodes[curNode].sons.begin(); for(; iter != nodes[curNode].sons.end(); iter++) { toId = *iter; if(!v[toId]) { v[toId] = true; int curNum = dfs(toId); curMax = maxv(curMax, curNum); total += curNum; } } total += 1; curMax = maxv(curMax, n - total); if(curMax < minBal) { minBal = curMax; minId = curNode; } return total; } int main() { int caseN, i, from, to; scanf("%d", &caseN); while(caseN--) { scanf("%d", &n); init(); for(i = 1; i < n; i++) { scanf("%d%d", &from, &to); nodes[from].sons.push_back(to); nodes[to].sons.push_back(from); } minBal = INT_MAX; dfs(1); printf("%d %d/n", minId, minBal); } return 0; }