poj_2367_Genealogical tree_很基础的拓扑排序

Genealogical tree
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 4292        Accepted: 2853        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(宗谱的) system leads to some embarrassment(窘迫).
 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(无子女的) assessors. However, the maintenance of this order really is not a trivial
  (不重要的) 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(后裔).

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 w(枚举)ith 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

很基础的拓扑排序，适合初学者练手

#include <stdio.h>#include <string.h>int g[105][105];int indegree[105];int s[105];int n,num;void topo_sort(){int i,j,k;for(i=1;i<=n;i++){for(j=1;j<=n;j++){if(indegree[j]==0){s[num++] = j;indegree[j] = -1;for(k=1;k<=n;k++){if(g[j][k]==1){indegree[k]--;}}break;}}}}int main(){while(scanf("%d",&n)!=EOF){int i,j,x;memset(g,0,sizeof(g));memset(indegree,0,sizeof(indegree));for(i=1;i<=n;i++){while(scanf("%d",&x) && x){g[i][x] = 1;indegree[x] ++ ;}}num = 1;topo_sort();for(i=1;i<=n;i++){printf("%d ",s[i]);}printf("\n");}return 0;}