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
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.
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 1Sample Output
1 2
题目分析 ： 找到树的重心，并输出序最小的那个，同时输出此时子树的最大结点数
代码示例 ：
const int maxn = 2e4+5;
#define ll long long
int n;
vector<int>edge[maxn];
int size[maxn], mx[maxn];
int root, balance;
void dfs(int x, int fa){
size[x] = 1;
mx[x] = 0;
for(int i = 0; i < edge[x].size(); i++){
int to = edge[x][i];
if (to == fa) continue;
dfs(to, x);
size[x] += size[to];
mx[x] = max(mx[x], size[to]);
}
mx[x] = max(mx[x], n-size[x]);
if (mx[x] < balance){
balance = mx[x];
root = x;
}
//else if (mx[x] == balance){
//root = min(root, x);
//}
}
int main() {
int t;
int a, b;
cin >> t;
while(t--){
cin >> n;
for(int i = 1; i <= n; i++){
edge[i].clear();
}
for(int i = 1; i < n; i++) {
scanf("%d%d", &a, &b);
edge[a].push_back(b);
edge[b].push_back(a);
}
balance = 99999999;
dfs(1, 1);
//dfs2(1, 1);
//for(int i = 1; i <= n; i++){
//cout << mx[i] << " " ;
//}
printf("%d %d\n", root, balance);
}
return 0;
}
