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10305 - Ordering Tasks

John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed.

Input

The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 andm. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0will finish the input.

Output

For each instance, print a line with n integers representing the tasks in a possible order of execution.

Sample Input

5 4

1 2

2 3

1 3

1 5

0 0

Sample Output

1 4 2 5 3

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=110;
int mp[maxn][maxn];
int toposort[maxn];
int vis[maxn];
int n,m,cnt;
void dfs(int x)
{
    vis[x]=1;
    for(int i=1;i<=n;i++)
        if(!vis[i]&&mp[x][i]&&i!=x)   dfs(i);
    toposort[cnt++]=x;
}
void solve()
{
    CLEAR(vis,0);
    for(int i=1;i<=n;i++)
       if(!vis[i])  dfs(i);
}
int main()
{
     int x,y;
     std::ios::sync_with_stdio(false);
     while(cin>>n>>m&&(n+m))
     {
         CLEAR(mp,0);
         while(m--)
         {
             cin>>x>>y;
             mp[x][y]=1;
         }
         cnt=0;
         solve();
         for(int i=cnt-1;i>=0;i--)
            printf(i==0?"%d\n":"%d ",toposort[i]);
     }
     return 0;
}

确定比赛名次

4 3
1 2
2 3
4 3

1 2 4 3

字典序最小的。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=550;
int mp[maxn][maxn],vis[maxn];
int in[maxn],que[maxn];
int n,m,k;
void toposort()
{
    k=0;int i;
    while(k<n)
    {
        for(i=1;i<=n;i++)
        {
            if(!vis[i]&&!in[i])
            {
                que[++k]=i;
                vis[i]=1;
                break;
            }
        }
        for(int j=1;j<=n;j++)
        {
            if(mp[i][j])  --in[j];
        }
    }
}
void output()
{
    for(int i=1;i<=k;i++)
       printf(i==k?"%d\n":"%d ",que[i]);
}
int main()
{
     int x,y;
     std::ios::sync_with_stdio(false);
     while(~scanf("%d%d",&n,&m))
     {
         CLEAR(in,0);
         CLEAR(mp,0);
         CLEAR(vis,0);
         while(m--)
         {
             scanf("%d%d",&x,&y);
             if(!mp[x][y])
             {
               mp[x][y]=1;
               in[y]++;
             }
         }
         toposort();
         output();
     }
     return 0;
}

Legal or Not

3 2
0 1
1 2
2 2
0 1
1 0
0 0

YES
NO

判断能不能拓扑排序。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=110;
int mp[maxn][maxn],vis[maxn];
int in[maxn];
int n,m,k;
int toposort()
{
    k=0;int i;
    for(int t=0;t<n;t++)
    {
      for(i=0;i<n;i++)
      {
        if(!vis[i]&&!in[i])
        {
            vis[i]=1;
            in[i]--;
            k++;
            for(int j=0;j<n;j++)
                if(mp[i][j])  in[j]--;
            break;
        }
      }
    }
//    cout<<"2333  "<<k<<endl;
    return k==n?1:0;
}
int main()
{
     int x,y;
     std::ios::sync_with_stdio(false);
     while(cin>>n>>m&&(n+m))
     {
         CLEAR(mp,0);
         CLEAR(vis,0);
         CLEAR(in,0);
         while(m--)
         {
             cin>>x>>y;
             if(!mp[x][y])
             {
                 mp[x][y]=1;
                 in[y]++;
             }
         }
         if(toposort())   cout<<"YES"<<endl;
         else   cout<<"NO"<<endl;
     }
     return 0;
}

拓扑排序入门

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原文地址：http://blog.csdn.net/u013582254/article/details/41710163