poj 2367 Genealogical tree

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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

504 5 1 01 05 3 03 0

Sample Output

2 4 5 3 1

这个题用是基础拓扑排序~有两种方法~

一种是按照算法导论上面的用时间戳~

一种是构图去计算入度为0的点~

我的代码大致类似于第一种~但因为是做完了题才去看的算法导论~所以又有所不同~

改天要再正式写一次~

关于拓扑排序~这篇文章介绍得挺详细~

http://hi.baidu.com/rain_bow_joy/blog/item/28daec17a049634020a4e9fd.html

这几篇是关于这道题的~

http://hi.baidu.com/%B5%AD%B5%C4%B2%E8/blog/item/4f388b0f952636c77acbe164.html

http://hi.baidu.com/just_need_you_/blog/item/2b33e031115e2d86a61e12a7.html

我的想法是设置一个二维数组~初始化为全1~输入时有数字则为0~

输入完后~数组每列和每行1的个数分别代表入度和出度~

若某a[i][x]一列全为1~则说明入度为0~输出该数~

然后将该列标记为全0~

a[x][i]这一行标记为全1~

再次计算~循环n次~

下面是代码~

#include"stdio.h"
#include"string.h"
int main()
{
int a[150][150];
int n;
int i,j;
int l;
int w;
int t=0;

freopen("01in.txt","r",stdin);
freopen("01out.txt","w",stdout);
scanf("%d",&n);
memset(a,'1',sizeof(a));

for(i=1;i<=n;i++)
  for(;;)
  {
   scanf("%d",&l);
   if(l==0)
    break;
   a[i][l]=0;
  }

  while(t<n)
  {
   for(i=1;i<=n;i++)
   {
    w=0;
    for(j=1;j<=n;j++)
    {
     if(a[j][i]==0)
      w=1;
    }
    if(w==0)
    {
     for(j=1;j<=n;j++)
      a[j][i]=0;
     t++;
     printf("%d ",i);
     for(j=1;j<=n;j++)
     a[i][j]=1;
     break;

    }
   }
  }


  return 0;

}