UVA 10305 Ordering Tasks（拓扑排序入门）【刘汝佳算法入门经典例6-15】

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 and m. 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 = 0 will 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

题意：有n个任务，每个任务与 其他任务有对应的小于关系，给出了m个关系，让你对这些任务进行排序，输出其中的一个序列。
现在没有字典序的要求，我们只需对其进行dfs就好了。
下面试AC代码：

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;const int MAXN=200;int n,m,map[MAXN][MAXN];int c[MAXN];int topo[MAXN],t;bool dfs(int u){   c[u]=-1;//访问标志   for(int v=1;v<=n;v++)   {      if(map[u][v])      {         if(c[v]<0)//判断是否为有向环         {            return false ;         }         else if (!c[v])         {         dfs(v) ;         }       }    }    c[u] = 1 ;    topo[--t] = u ;//至于这里为什么这样子写，大家可以思考一下    return true ;}bool toposort(){    t=n;    memset(c,0,sizeof(c));    for(int i=1;i<=n;i++)    {       if(!c[i])       {           if(!dfs(i))           {            return false ;           }       }     }    return true ;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        if(m==0&&n==0) break;        memset(map,0,sizeof(map));        memset(topo,0,sizeof(topo));        for(int i=0;i<m;i++)        {            int u,v;            scanf("%d%d",&u,&v);            map[u][v]=1;        }        toposort();        for(int i=0;i<=n-2;i++)        printf("%d ",topo[i]);        printf("%d\n",topo[n-1]);    }    return 0;}

下面是另外一种做法，不过下面这种做法输出的结果是按着字典序从小到大的要求。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int n,m,vis[1005],e[1005][1005];bool judge(int k)//判断还有没有比k更小的{    for(int i=1; i<=n; i++)        if(e[k][i])        {            return 0;        }    return 1;}bool clea(int k)//输出k以后，比k小的关系就不需要管了{    for(int i=1; i<=n; i++)    {        e[i][k]=0;    }}int main(){    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0)            break;        memset(e,0,sizeof(e));        int x,y;        memset(vis,0,sizeof(vis));        for(int i=0; i<m; i++)        {            scanf("%d%d",&x,&y);            e[y][x]=1;        }        int cnt=0;        while(cnt<n)        {            for(int i=1; i<=n; i++)            {                if(vis[i])                    continue;                if(judge(i))                {                    if(cnt)                        printf(" ");                    printf("%d",i);                    vis[i]=1;                    cnt++;                    clea(i);                    break;                }            }        }        printf("\n");    }    return 0;}