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.
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 Output
2 4 5 3 1
拓扑排序 输入一个n
然后接下来的第n行输出排在n后面的数字序列 输出0时此行结束
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
using namespace std;
int map[1001][1001];
int die[1001];
int cun[1001];
void topo(int n)
{
int i,j,first,cut=0;
for(i=0;i<n;i++)
{
for(j=n;j>0;j--)
{
if(die[j]==0)
{
first=j;
break;
}
}
cun[cut++]=first; die[first]=-1;
for(j=1;j<=n;j++)
{
if(map[first][j]==1)
{
die[j]--;
}
}
}
printf("%d",cun[0]);
for(i=1;i<n;i++)
{
printf(" %d",cun[i]);
}
printf("\n");
}
编程五分钟,调试两小时...