分析

题目要求求出当前点子树dfs遍历时第v个到达的点的编号
那么我们可以直接走dfn序
然后用一个数组记录
那么每次只需要特判加输出即可
时间复杂度：
期望得分：100分

代码

//CF1006E
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

#define ll long long
#define cl(x, y) memset((x), (y), sizeof(x))
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define per(i, a, b) for(int i = a; i >= b; i--)
#define de(x) cerr << #x << " = " << x << " "
#define inc_mod(x, y) x = ((x - y) % mod + mod) % mod
#define add_mod(x, y) x = (x + y) % mod
#define lowbit(x) (x & (-x))
#define inf 0x3f3f3f3f
#define mod 998244353
#define rson (x << 1 | 1)
#define lson (x << 1)
using namespace std;
const int maxn = 2e5 + 10;

int nxt[maxn << 1], ed[maxn << 1], head[maxn], cur;
int dfn[maxn], dfn_num, dot[maxn], size[maxn];
int total, u, v, w, opt[maxn], test;

inline void file() {
    freopen(".in", "r", stdin);
    freopen(".out", "w", stdout);
}

inline void dfs(int now, int pre) {
    dfn[now] = ++dfn_num;
    dot[dfn_num] = now;
    size[now] = 1;
    for(int i = head[now]; i; i = nxt[i]) {
        if(ed[i] == pre) continue;;
        dfs(ed[i], now);
        size[now] += size[ed[i]];
    }
}

inline void add_edge(int from, int to) {
    nxt[++cur] = head[from];
    head[from] = cur;
    ed[cur] = to;
} 

int main() {
    // file();
    ios::sync_with_stdio(false);
    cin >> total >> test;
    rep(i, 2, total) cin >> opt[i];
    per(i, total, 2) { add_edge(opt[i], i); }
    dfs(1, 0);
    while(test--) {
        cin >> u >> v;
        if(size[u] >= v) { cout << dot[dfn[u] + v - 1] << endl; }
        else { cout << "-1" << endl; }
    }
    // fclose(stdin);
    // fclose(stdout);
    system("pause");
    return 0;
}