Balancing Act

時限：1000ms

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

樹的重心也叫樹的質心。找到一個點,其所有的子樹中最大的子樹節(jié)點數最少,那么這個點就是這棵樹的重心,刪去重心后，生成的多棵樹盡可能平衡。
利用樹形dp的思想，對節(jié)點u的子樹進行遍歷。和樹的直徑不同的是，如果子樹的分割確定了，那么子樹之外的節(jié)點必定在一顆樹上，不需要在dfs一次。
#include "stdio.h" #include "algorithm" #include "string.h" using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 20000 + 10; bool vis[maxn]; struct edge {int to, next; } e[maxn*2]; int n, num[maxn]; int size, cur; int tot = 0; int head[maxn]; void add_edge(int u, int v) {e[tot].to = v; e[tot].next = head[u];head[u] = tot++; } void dfs(int u) {num[u] = 0;int tr = 0;vis[u] = true;for (int i = head[u]; i != -1; i = e[i].next) {int v = e[i].to;if (vis[v]) continue;dfs(v);num[u] += num[v] + 1;tr = max(num[v] + 1, tr);} tr = max(n - num[u] - 1, tr); //剩下那棵樹的大小if (tr < size || tr == size && cur > u) { //答案還要編號節(jié)點最小size = tr; cur = u;} } void init() {memset(vis, false, sizeof(vis));memset(head, -1, sizeof(head));tot = 0; size = INF; } int main(int argc, char const *argv[]) {int T;scanf("%d", &T);while (T--) {init();scanf("%d", &n);for (int i = 0; i < n-1; i++) {int u, v;scanf("%d%d", &u, &v); add_edge(u, v);add_edge(v, u);}dfs(1);printf("%d %d\n", cur, size);}return 0; }

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轉載于:https://www.cnblogs.com/cniwoq/p/7247542.html

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