Once upon a time there was only one router in the well-known company Bmail. Years went by and over time new routers were purchased. Every time they bought a new router, they connected it to one of the routers bought before it. You are given the values pi — the index of the router to which the i -th router was connected after being purchased (pi<i ).
There are n routers in Boogle in total now. Print the sequence of routers on the path from the first to the n -th router.
Input
The first line contains integer number n (2≤n≤200000 ) — the number of the routers. The following line contains n−1 integers p2,p3,…,pn (1≤pi<i ), where pi is equal to index of the router to which the i-th was connected after purchase.
Output
Print the path from the 1-st to the n-th router. It starts with 1 and ends with n. All the elements in the path should be distinct.
Examples
Input
8
1 1 2 2 3 2 5
Output
1 2 5 8
Input
6
1 2 3 4 5
Output
1 2 3 4 5 6
Input
7
1 1 2 3 4 3
Output
1 3 7
题目分析，就是找父亲节点，我是从最后一个点开始找然后一直到第一个开头的节点然后结束循环，具体实现看代码。
听懂了记得给个赞鼓励一下，码字不易，用爱发电。
上ac代码。
有事你就q我；QQ2917366383
学习算法
#include <bits/stdc++.h> using namespace std; int n,m,i,j,k,fa[1000001],t,qwq,b[1000001]; int main(){ scanf("%d",&n); for (i=2;i<=n;i++) scanf("%d",&fa[i]); j=0;qwq=n;//这里n的意思就是开始从最后一个点 fa[1]=0; while (fa[qwq]!=0){//这里是找父亲节点 j++; b[j]=qwq; qwq=fa[qwq]; //这里相当于连接下一个节点 } cout<<1<<" "; for (i=j;i>=1;i--) cout<<b[i]<<" "; return 0; }
