poj2367Genealogical tree 拓扑排序

Genealogical tree

 

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

思路：事实证明，有个读题的同桌就是爽，直接讲大意，读题？tan90（坏笑）

直接看样例，第一行一个整数n代表共有从1到n的n个数，随后n行第i行的多个整数代表这i个整数的次数要排在i后面，以0代表此行输入结束

现在来看思路，很简单，建图拓扑排序就ok了，你可能还会疑问怎么建图，很简单，就是暴力对于让第i行的数都指向i

代码：

#include<cstdio>
#include<cstring>
#define mem(a) sizeof(a,0,sizeof(a))
using namespace std;
const int maxn = (int)1e3 + 10;
int G[maxn][maxn];
int vis[maxn];
int n;
void toposort()
{
	for (int i = 1;i <= n;i ++)
	{
		for (int j = 1;j <= n;j ++)
		{
			if (!vis[j])
			{
				printf("%d",j);
				vis[j] --;
				if (i != n) putchar(' ');
				else putchar('\n');
				for (int k = 1;k <= n;k ++)
				{
					if (G[j][k])
						vis[k]--;
				}
				break;
			}
		}
	}
}
int main()
{
	scanf("%d",&n);
	for (int i = 1;i <= n;i ++)
	{
		int x;
		while (scanf("%d",&x) && x)
		{
			if (!G[i][x])
			{
				G[i][x] = 1;
				vis[x] ++;
			}
		}
	}
	toposort();
	return 0;
}