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。
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#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 () {
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) {
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;
}
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