Military Problem

题目地址：

https://ac.nowcoder.com/acm/problem/112932

基本思路：

题目很长,可以概况为给你一棵树,每次查询,
你要找到从开始按照遍历顺序,遍历到的第个节点,
也就是找到点对应序后第个的节点就是了,
然后注意超出子树的范围了要输出。

参考代码：

#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define ll long long
#define ull unsigned long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define debug(x) cerr << #x << " = " << x << '\n';
#define pll pair <ll, ll>
#define fir first
#define sec second
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 2e5 + 10;
int n,m;
vector<int> G[maxn];
int dfn[maxn],id[maxn],mx[maxn];
signed main() {
  n = read(),m = read();
  rep(i,2,n){
    int p = read();
    G[i].push_back(p);
    G[p].push_back(i);
  }
  int tot = 0;
  function<void(int,int)> dfs = [&](int u,int par){
    dfn[u] = ++tot;
    id[tot] = u;
    for(auto to : G[u]){
      if(to == par) continue;
      dfs(to,u);
    }
    mx[u] = tot;
  };
  dfs(1,0);
  for(int i = 1 ; i <= m ; i++){
    int u = read(),k = read();
    int now = dfn[u] + k - 1;
    int res = -1;
    if(now <= mx[u]) res = id[now];
    print(res); puts("");
  }
  return 0;
}