poj2367Genealogical tree

poj2367Genealogical tree

題目鏈接：

點我點我

題目：

Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 2704 Accepted: 1816 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 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

Source

Ural State University Internal Contest October'2000 Junior Session

這個題目是拓撲排序的入門題。。

首先在這裡一個講的非常好的鏈接：

傳送門點我點我

我用了兩種方法做這個題。

第一種是利用入度為0的點必然是前面的點，然後刪除從這個點到其他點的邊，最後一期輸出結果。。速度很快。

第二種是利用dfs搜索，直到搜索到已經訪問到的點，然後利用棧來保存。。最後利用棧的性質來輸出即可。。

第二中代碼為：

#include
#include
#include
#include
#include
using namespace std;
const int maxn=100+10;
int vis[maxn],map[maxn][maxn];
int n,t;
stackS;
bool dfs(int u)
{
    vis[u]=-1;
    for(int v=1;v<=n;v++)
     if(map[u][v])
    {
       // if(vis[v]<0)  return false;
         if(!vis[v]&&!dfs(v))  return false;
    }
    vis[u]=1;
    S.push(u);
    return true;
}


bool toposort()
{
    memset(vis,0,sizeof(vis));
    for(int u=1;u<=n;u++)
      {
        if(!vis[u])
        {
          if(!dfs(u))
            return false;
        }
     }
     return true;
}


int main()
{
    int u;
    bool ok;
    while(scanf("%d",&n)!=EOF)
    {
        t=n;
        memset(map,0,sizeof(map));
        for(int i=1;i<=n;i++)
            {
                while(1)
                {
                    scanf("%d",&u);
                    if(u==0)
                        break;
                    map[i][u]=1;
                }
            }
        ok=toposort();
        if(ok)
        {
        while(!S.empty())
           {
               int val=S.top();
               if(t!=1)
                  printf("%d ",val);
               else
                  printf("%d\n",val);
               S.pop();
               t--;
           }
        }
    }
    return 0;
}

第一種方法代碼：

#include
#include
const int maxn=100+10;
int map[maxn][maxn],into[maxn],ans[maxn],vis[maxn];
int pos;
int main()
{
    int n,u,temp;
    while(~scanf("%d",&n))
    {
        pos=0;
        memset(map,0,sizeof(map));
        memset(into,0,sizeof(into));
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            while(1)
            {
                scanf("%d",&u);
                if(u==0)
                  break;
                map[i][u]=1;
            }
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
        {
            if(map[i][j])
                into[j]++;
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
        {
            if(into[j]==0&&!vis[j])
              {
                temp=j;
                vis[j]=1;
                ans[pos++]=temp;
                for(int m=1;m<=n;m++)
                 {
                   if(map[temp][m])
                       into[m]--;
                 }
              }
        }
        for(int i=0;i