It's the same as the given sample input.

In addition I get WA6 and this is my code :
/*
#include <stdio.h>
#define maxn 105

int vp[maxn], pv[maxn];

int main() {

    int n, x, y;
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++) {
        vp[i] = i;
        pv[i] = i;
    }
    for (int i = 1; i <= n; i ++) {
        int min = vp[i];
        while(1) {
            scanf("%d", &x);
            if(x == 0) break;
            if(vp[x] < min) min = vp[x];
        }
        for (int j = i - 1; j >= min; j --) {
            pv[j + 1] = pv[j];
            vp[pv[j + 1]] = j + 1;
        }
        pv[min] = i;
        vp[i] = min;
    }
    for (int i = 1; i <= n; i ++) {
        printf("%d ", pv[i]);
    }
    return 0;
}
*/

My AC program gives "3 2 1".
So, probably "3 2 1"  is the right answer.

If several sequences satisfy the conditions of the problem, you are to write to the standard output any of them!!! ;) ;/

I think the problem is that a children could have more than 1 parent - and here you have to check if all of his parents have been printed. If you can't find the solution yourself - write and I will tell you a way.

P.S. Other people say that "One of the lines in test 4 has trailing space. This old problem was designed for Pascal/C++, such inputs treated as a bad style nowadays."
I don't know what this means, but first check if this is the problem!

Edited by author 07.01.2015 04:12

This is my code. Can someone tell me why do I get WA#4?

import java.util.Collection;
import java.util.Comparator;
import java.util.HashSet;
import java.util.Hashtable;
import java.util.Iterator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class MartianCouncil {

    public static void main(String[] args) {
        Hashtable<Integer, TreeNode> global = new Hashtable<Integer, TreeNode>();
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        if (n == 0) {
            s.close();
            return;
        }
        s.nextLine();
        for (int i = 1; i <= n; i++) {
            TreeNode node = global.get(i);
            if (node == null) {
                node = new TreeNode(i);
                global.put(i, node);
            }
            String line = s.nextLine();
            String[] nums = line.split(" ");
            for (int j = 0; j < nums.length - 1; j++) {
                int num = Integer.parseInt(nums[j]);
                TreeNode nd = global.get(num);
                if (nd == null) {
                    nd = new TreeNode(num);
                    global.put(num, nd);
                }
                node.addChild(nd);
            }
        }
        s.close();
        print(global);
    }

    public static void print(Hashtable<Integer, TreeNode> global) {
        PriorityQueue<TreeNode> q = new PriorityQueue<TreeNode>();
        q.addAll(global.values());
        StringBuilder res = new StringBuilder("");
        while (!q.isEmpty()) {
            TreeNode nd = q.poll();
            res = res.append((res.length() == 0) ? "" : " ");
            res = res.append(nd.getNum());
        }
        System.out.print(res);
    }


    public static class TreeNode implements Comparable<MartianCouncil.TreeNode> {

        private int num = 0;
        private Collection<MartianCouncil.TreeNode> children = new HashSet<MartianCouncil.TreeNode>();
        private Collection<MartianCouncil.TreeNode> parents = new HashSet<MartianCouncil.TreeNode>();

        public TreeNode(int num) {
            this.num = num;
        }

        public int hashCode() {
            return num;
        }

        public boolean equals(Object o) {
            if (o == null) {
                return false;
            } else if (!(o instanceof MartianCouncil.TreeNode)) {
                return false;
            } else {
                return (this.hashCode() == o.hashCode());
            }
        }

        public boolean addChild(MartianCouncil.TreeNode child) {
            child.parents.add(this);
            return children.add(child);
        }

        public boolean removeChild(MartianCouncil.TreeNode child) {
            child.parents.remove(this);
            return children.remove(child);
        }

        public boolean addParent(TreeNode parent) {
            parent.children.add(this);
            return parents.add(parent);
        }

        public boolean removeParent(TreeNode parent) {
            parent.children.remove(this);
            return parents.remove(parent);
        }

        public void removeAllChildren() {
            Iterator<TreeNode> it = children.iterator();
            while (it.hasNext()) {
                TreeNode child = it.next();
                child.parents.remove(this);
            }
            children.clear();
        }

        public int getNum() {
            return num;
        }

        public Collection<TreeNode> getChildren() {
            return children;
        }

        public int getParentsNum() {
            return parents.size();
        }

        @Override
        public int compareTo(MartianCouncil.TreeNode nd) {
            return Math.round(Math.signum(this.getParentsNum() - nd.getParentsNum()));
        }

    }


    public class TreeNodeComp implements Comparator<MartianCouncil.TreeNode> {

        @Override
        public int compare(MartianCouncil.TreeNode nd1, MartianCouncil.TreeNode nd2) {
            return Math.round(Math.signum(nd1.getParentsNum() - nd2.getParentsNum()));
        }

    }

}

DFS doesn't work here..

One of the lines in test 4 has trailing space. This old problem was designed for Pascal/C++, such inputs treated as a bad style nowadays.

Hey
Probably because N <= 100 :-D

Is it possible to determine inputs? Are there exist resource&? In the 4-th test I have problem time limit exception.

chunki 3 5ni farzandi

Edited by author 08.07.2011 16:39

Because 5 is parent of 3

the graph may be not connected.

I got WA#2 too,
but I thought that the graph is supposed to be not connected.

Can help me identify my errors , here's my solution :
- I reverse all the edges
- DFS from the vertex without any other vertex pointing to
- Count the depth, if depth = number of vertex -> output in reverse order

3Q,i made the same mistake,and i now get AC

2170208 13:57:22 23 июл 2008 Mansurov Artur 1022 C++ Accepted 0.001 153 КБ

O(n^2) simple search