timus 1022 Genealogical Tree(拓扑排序)

Genealogical Tree

Time limit: 1.0 second
Memory limit: 64 MB

Background

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.

Problem

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 integers 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

5
0
4 5 1 0
1 0
5 3 0
3 0

2 4 5 3 1

Problem Author: Leonid Volkov

【分析】英文渣渣，一开始看了半天都不知道在说什么，后来才知道题意。是这样的，火星人血统非常乱，乱到无法理解，好像是开会的时候，要把议员席给祖先，

也就是说每个人都得在自己的长辈都坐下之后才能坐下。很明显就是个拓扑排序。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define inf 10000000
#define mod 10000
typedef long long ll;
using namespace std;
const int N=605;
const int M=50000;
int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
int in[N],vis[N];
int n,m,k;
vector<int>vec[N];
void dfs(int x)
{
    printf("%d ",x);
    int f=0;
    vis[x]=1;
    for(int i=0;i<vec[x].size();i++){
        int v=vec[x][i];
        if(in[v])in[v]--;
    }
    for(int i=1;i<=n;i++)if(!in[i]&&!vis[i])f=1,dfs(i);
    if(!f) return;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        while(~scanf("%d",&m)&&m){
            in[m]++;
            vec[i].push_back(m);
        }
    }
    for(int i=1;i<=n;i++){
        if(!in[i]&&!vis[i]){
            dfs(i);
        }
    }
    printf("\n");
    return 0;
}