Codeforces Round #498 (Div. 3) E. Military Problem

http://codeforces.com/contest/1006/problem/E

E. Military Problem

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In this problem you will have to help Berland army with organizing their command delivery system.

There are $\mathrm{nn officers\; in\; Berland\; army.\; The\; first\; officer\; is\; the\; commander\; of\; the\; army,\; and\; he\; does\; not\; have\; any\; superiors.\; Every\; other\; officer\; has\; exactly\; one\; direct\; superior.\; If\; officer aa is\; the\; direct\; superior\; of\; officer bb,\; then\; we\; also\; can\; say\; that\; officer bb is\; a\; direct\; subordinate\; of\; officer aa.}$

Officer $\mathrm{xx is\; considered\; to\; be\; a\; subordinate\; (direct\; or\; indirect)\; of\; officer yy if\; one\; of\; the\; following\; conditions\; holds:}$

• officer $\mathrm{yy is\; the\; direct\; superior\; of\; officer xx;}$
• the direct superior of officer $\mathrm{xx is\; a\; subordinate\; of\; officer yy.}$

For example, on the picture below the subordinates of the officer $\mathrm{33 are: 5,6,7,8,95,6,7,8,9.}$

The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.

Formally, let's represent Berland army as a tree consisting of $\mathrm{nn vertices,\; in\; which\; vertex uu corresponds\; to\; officer uu.\; The\; parent\; of\; vertex uucorresponds\; to\; the\; direct\; superior\; of\; officer uu.\; The\; root\; (which\; has\; index 11)\; corresponds\; to\; the\; commander\; of\; the\; army.}$

Berland War Ministry has ordered you to give answers on $\mathrm{qq queries,\; the ii-th\; query\; is\; given\; as (ui,ki)(ui,ki),\; where uiui is\; some\; officer,\; and kiki is\; a\; positive\; integer.}$

To process the $\mathrm{ii-th\; query\; imagine\; how\; a\; command\; from uiui spreads\; to\; the\; subordinates\; of uiui.\; Typical\; DFS\; (depth\; first\; search)\; algorithm\; is\; used\; here.}$

Suppose the current officer is $\mathrm{aa and\; he\; spreads\; a\; command.\; Officer aa chooses bb —\; one\; of\; his\; direct\; subordinates\; (i.e.\; a\; child\; in\; the\; tree)\; who\; has\; not\; received\; this\; command\; yet.\; If\; there\; are\; many\; such\; direct\; subordinates,\; then aa chooses\; the\; one\; having\; minimal\; index.\; Officer aa gives\; a\; command\; to\; officer bb.\; Afterwards, bb uses\; exactly\; the\; same\; algorithm\; to\; spread\; the\; command\; to\; its\; subtree.\; After bb finishes\; spreading\; the\; command,\; officer aa chooses\; the\; next\; direct\; subordinate\; again\; (using\; the\; same\; strategy).\; When\; officer aa cannot\; choose\; any\; direct\; subordinate\; who\; still\; hasn\text{'}t\; received\; this\; command,\; officer aa finishes\; spreading\; the\; command.}$

Let's look at the following example:

If officer $\mathrm{11 spreads\; a\; command,\; officers\; receive\; it\; in\; the\; following\; order: [1,2,3,5,6,8,7,9,4][1,2,3,5,6,8,7,9,4].}$

If officer $\mathrm{33 spreads\; a\; command,\; officers\; receive\; it\; in\; the\; following\; order: [3,5,6,8,7,9][3,5,6,8,7,9].}$

If officer $\mathrm{77 spreads\; a\; command,\; officers\; receive\; it\; in\; the\; following\; order: [7,9][7,9].}$

If officer $\mathrm{99 spreads\; a\; command,\; officers\; receive\; it\; in\; the\; following\; order: [9][9].}$

To answer the $\mathrm{ii-th\; query (ui,ki)(ui,ki),\; construct\; a\; sequence\; which\; describes\; the\; order\; in\; which\; officers\; will\; receive\; the\; command\; if\; the uiui-th\; officer\; spreads\; it.\; Return\; the kiki-th\; element\; of\; the\; constructed\; list\; or -1 if\; there\; are\; fewer\; than kiki elements\; in\; it.}$

You should process queries independently. A query doesn't affect the following queries.

Input

The first line of the input contains two integers $\mathrm{nn and qq (2\le n\le 2\cdot 105,1\le q\le 2\cdot 1052\le n\le 2\cdot 105,1\le q\le 2\cdot 105)\; —\; the\; number\; of\; officers\; in\; Berland\; army\; and\; the\; number\; of\; queries.}$

The second line of the input contains $\mathrm{n-1n-1 integers p2,p3,\dots ,pnp2,p3,\dots ,pn (1\le pi<i1\le pi<i),\; where pipi is\; the\; index\; of\; the\; direct\; superior\; of\; the\; officer\; having\; the\; index ii.\; The\; commander\; has\; index 11 and\; doesn\text{'}t\; have\; any\; superiors.}$

The next $\mathrm{qq lines\; describe\; the\; queries.\; The ii-th\; query\; is\; given\; as\; a\; pair\; (ui,kiui,ki)\; (1\le ui,ki\le n1\le ui,ki\le n),\; where uiui is\; the\; index\; of\; the\; officer\; which\; starts\; spreading\; a\; command,\; and kiki is\; the\; index\; of\; the\; required\; officer\; in\; the\; command\; spreading\; sequence.}$

Output

Print $\mathrm{qq numbers,\; where\; the ii-th\; number\; is\; the\; officer\; at\; the\; position kiki in\; the\; list\; which\; describes\; the\; order\; in\; which\; officers\; will\; receive\; the\; command\; if\; it\; starts\; spreading\; from\; officer uiui.\; Print\; "-1"\; if\; the\; number\; of\; officers\; which\; receive\; the\; command\; is\; less\; than kiki.}$

You should process queries independently. They do not affect each other.

Example

input

Copy

9 6
1 1 1 3 5 3 5 7
3 1
1 5
3 4
7 3
1 8
1 9

output

Copy

3
6
8
-1
9
4
题意：给你n个节点的数和q次查询
节点输入规则是i的父节点是di
查询规则是，求从节点a开始，查询他的子树的第i个节点
题解：先用dfs预处理每一个节点对应的子树的数量和该数量对应的节点个数
代码如下

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

const int mx = 2e5+10;
vector<int> v[mx];
int vis[mx], sum[mx], cnt = 1, s[mx];
//预处理dfs序
int dfs(int id){
    int ans = 1;
    vis[id] = cnt;  //其子树的数量
    s[cnt] = id;   //对应子树数量对应的节点
    cnt++;
    for (int i = 0; i < v[id].size(); i++){
        ans += dfs(v[id][i]);   //访问节点
    }
    return sum[id] += ans;  //返回子树个数
}

int main(){
    int n, q, p, x, y;
    scanf("%d%d", &n, &q);
    for (int i = 1; i < n; i++){
        scanf("%d", &p);
        v[p].push_back(i+1);//子节点
    }
    dfs(1);
    while (q--){
        scanf("%d%d", &x, &y);
        if (sum[x] < y) printf("-1\n");
        else printf("%d\n", s[vis[x]+y-1]);
    }
    return 0;
}