CCF 201503-04 网络延时

遍历树，获取每个节点的深度。对每个节点，考虑深度差最大的两个子树（即获取到一个较大时延），以及子树中最大的时延，比较得出最大。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 20015;
vector<int> tree[N];
#define getMax2Deep(a,b,c) (c>a?(b=a,a=c):((c>b)?(b=c):(1)))
int deep[N] = { 0 };
int n, m;
void buildTree() {
	int i = 2;
	int root;
	fill(deep, deep + N, 0);
	while (i<=n) {
		cin >> root;
		tree[root].push_back(i);
		i++;
	}
	while (i<=n + m) {
		cin >> root;
		tree[root].push_back(i);
		i++;
	}
}
int dfs(int root = 1) {//return max delay in this sub-tree
	int maxDelay = 0;
	int maxDeep = 0, max2deep = 0;
	if (tree[root].size() == 0) {
		deep[root] = 1;
		return 1;
	}
	for (int i = 0; i<tree[root].size(); i++) {
		maxDelay = max(maxDelay, dfs(tree[root][i]));
		getMax2Deep(maxDeep, max2deep, deep[tree[root][i]]);
	}
	deep[root] = maxDeep + 1;
	return max(maxDelay, maxDeep+max2deep);
}
int main() {
	freopen("in.txt", "r", stdin);
	cin >> n >> m;
	buildTree();
	cout << dfs();
	return 0;
}