POJ2367——Genealogical tree

Description

The system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They gather together in different groups, so that a Martian can have one parent as well as ten. Nobody will be surprised by a hundred of children. Martians have got used to this and their style of life seems to them natural.
And in the Planetary Council the confusing genealogical system leads to some embarrassment. There meet the worthiest of Martians, and therefore in order to offend nobody in all of the discussions it is used first to give the floor to the old Martians, than to the younger ones and only than to the most young childless assessors. However, the maintenance of this order really is not a trivial task. Not always Martian knows all of his parents (and there's nothing to tell about his grandparents!). But if by a mistake first speak a grandson and only than his young appearing great-grandfather, this is a real scandal.
Your task is to write a program, which would define once and for all, an order that would guarantee that every member of the Council takes the floor earlier than each of his descendants.

Input

The first line of the standard input contains an only number N, 1 <= N <= 100 — a number of members of the Martian Planetary Council. According to the centuries-old tradition members of the Council are enumerated with the natural numbers from 1 up to N. Further, there are exactly N lines, moreover, the I-th line contains a list of I-th member's children. The list of children is a sequence of serial numbers of children in a arbitrary order separated by spaces. The list of children may be empty. The list (even if it is empty) ends with 0.

Output

The standard output should contain in its only line a sequence of speakers' numbers, separated by spaces. If several sequences satisfy the conditions of the problem, you are to write to the standard output any of them. At least one such sequence always exists.

Sample Input

5
0
4 5 1 0
1 0
5 3 0
3 0

Sample Output

2 4 5 3 1

Source

Ural State University Internal Contest October'2000 Junior Session


拓扑排序，算法本身很简单，建立一张DAG，然后用一个队列维护入度为0的点，处理完就把这些点和关联的边删除，把新的人度为0的点放入队列，直到队列为空，循环结束后，如果扫描到的点少于图中的点，说明有回路，否则就存在这样的序列。

当然本题保证至少有1个这样的序列。

#include<cstdio>
#include<string>
#include<queue>
#include<iostream>
#include<algorithm>

using namespace std;

const int maxn=105;

struct node
{
	int to;
	int next;
}edge[maxn*4];
int head[maxn];
int tot,n,cnt;
int in[maxn];
int sorted[maxn];

void addedge(int from,int to)
{
	edge[tot].to=to;
	edge[tot].next=head[from];
	head[from]=tot++;
}

void topo_sort()
{
	cnt=0;
	memset(in,0,sizeof(in));
	for(int i=1;i<=n;i++)
		for(int j=head[i];j!=-1;j=edge[j].next)
			in[edge[j].to]++;
	queue<int>qu;
	while(!qu.empty())
	  	qu.pop();
	for(int i=1;i<=n;i++)
		if(!in[i])
			qu.push(i);	
	while(!qu.empty())
	{
		int x=qu.front();
		qu.pop();
		sorted[cnt++]=x;
		for(int i=head[x];i!=-1;i=edge[i].next)
		{
			in[edge[i].to]--;
			if(!in[edge[i].to])
				qu.push(edge[i].to);
		}	
	}
	//if(cnt==n)
		//return 1;
//	return 0;
}

int main()
{
	int j;
	while(~scanf("%d",&n))
	{
		memset(head,-1,sizeof(head));
		tot=0;
		for(int i=1;i<=n;i++)
		{
			while(scanf("%d",&j) && j)
				addedge(i,j);
		}
		topo_sort();
		for(int i=0;i<cnt-1;i++)
			printf("%d ",sorted[i]);
		printf("%d\n",sorted[cnt-1]);
	}
	return 0;
}