DFS序模板题 CF498Div3E

题目链接: http://codeforces.com/contest/1006/problem/E

题意: 给定一棵树, 进行多次查询——查询某节点子树DFS序列的某一项

思路: 求出该树DFS序, 进行查询即可

总结: 第一道DFS序题目.

代码如下:

import java.util.*;

public class Main {
    static int tot;
    static ArrayList<Integer>[] g;
    static int[] in, out, dfs;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        in = new int[n + 1];
        out = new int[n + 1];
        dfs = new int[n + 1];
        g = new ArrayList[n + 1];
        for (int i = 1; i <= n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 2; i <= n; i++) {
            int t = sc.nextInt();
            g[i].add(t);
            g[t].add(i);
        }
        dfs(1, 0);
        for (int i = 0; i < m; i++) {
            int u = sc.nextInt(), k = sc.nextInt();
            k--;
            if (out[u] - in[u] >= k) {
                System.out.println(dfs[in[u] + k]);
            } else {
                System.out.println(-1);
            }
        }
    }

    public static void dfs(int u, int from) {
        in[u] = ++tot;
        dfs[tot] = u;
        for (int i = 0; i < g[u].size(); i++) {
            if (g[u].get(i) != from) {
                dfs(g[u].get(i), u);
            }
        }
        out[u] = tot;
    }

}