poj——2367 Genealogical tree

　　　　　　　　　　　　　　　　　　　　　　　　　　Genealogical tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6025		Accepted: 3969		Special Judge

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

题目重现：

火星人血缘关系的制度令人困惑。实际上，火星人要发芽，想要什么。他们聚集在一起不同的群体，所以一个火星人可以有一个父母以及十个。没有人会惊讶于一百个孩子。火星人习惯了，他们的生活方式似乎是自然的。而在行星委员会，混乱的系谱系统会导致一些尴尬。遇到最有价值的火星人，所以为了在所有的讨论中冒犯任何人，首先要让老火星人，而不是对年轻人，而不是最年轻的无孩子的陪审员。然而，维护这个命令真的不是一件小事。不总是火星人知道他所有的父母（没有什么可以告诉他的祖父母！）。但是，如果一个错误首先说出一个孙子，而不是他的年轻人出现的爷爷，这是一个真正的丑闻。你的任务是编写一个程序，一劳永逸地定义一个可以保证安理会每一位成员比每个后裔早日发言的命令

思路：拓扑排序模板、、、

代码：

#include<queue>
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 1010
using namespace std;
queue<int>q;
int n,m,s,tot,in[N],ans[N],head[N];
struct Edge
{
    int to,next,from;
}edge[N];
int add(int x,int y)
{
    tot++;
    edge[tot].to=y;
    edge[tot].next=head[x];
    head[x]=tot;
}
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
int main()
{
    n=read();
    ;i<=n;i++)
      )
      {
         m=read();
         ) break;
         in[m]++;add(i,m);
      }
    ;i<=n;i++)
     ) q.push(i);
    while(!q.empty())
    {
        m=q.front(),q.pop();
        ans[++s]=m;
        for(int i=head[m];i;i=edge[i].next)
        {
            int t=edge[i].to;
            in[t]--;
            ) q.push(t);
        }
    }
    ;i<=s;i++)
     printf("%d ",ans[i]);
    ;
}