读不懂题的树
A - TreeScript
vjudge上面看到的比较简洁的写法
据说是一种树形dp
转移方程就是父亲节点的寄存器数量等于max（最大子树的寄存器数量，次大子树的寄存器数量+1）
脑子转不过来不想转
//一种vector用法
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
vector<int> G[N];
int dfs(int u, int fa) {
if (G[u].size() == 0) {
return 0;
}
vector<int> vec;//
for (auto v : G[u]) {
if (v == fa)
continue;
vec.push_back(dfs(v, u));
}
sort(vec.begin(), vec.end());
//cout << "v:" << u << " fa:" << fa << endl;
//for (auto x : vec) {
// cout << x << " ";
//}
//cout << endl;
if (vec.size() == 1) {
return vec[0];
} else {
return max(vec[vec.size() - 2] + 1, vec[vec.size() - 1]);
}
}
void solve() {
int n, rt = 1;
cin >> n;
for (int i = 1; i <= n; i++) {
G[i].clear();
}
for (int i = 1, x; i <= n; i++) {
cin >> x;
if (x == 0) {
rt = i;
}
G[x].push_back(i);
}
cout << dfs(rt, 0) + 1 << endl;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int T;
cin >> T;
while (T--) {
solve();
}
return 0;
}