UVA 10305 Ordering Tasks

题目链接：UVA 10305 Ordering Tasks

Ordering Tasks

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

 

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

拓扑排序，基于dfs实现，其实就是记录dfs每个节点访问时间，然后逆序输出，兴趣读者可以参看算法导论图论章节。

#include 
#include 
#include 
using namespace std;

bool visited[110];
int g[110][110];
int n, m;
int topoSort[110];
int cnt;
void dfs(int s)
{
    visited[s] = true;
    for(int j = 1; j <= n; j++)
        if(j!= s && g[s][j]==1 && !visited[j])
            dfs(j);
    topoSort[cnt++] = s;
}
void dfs_travel()
{
    cnt = 0;
    memset(visited, false, sizeof(visited));
    for(int i = 1; i <= n; i++)
        if(!visited[i])
            dfs(i);
}
int main()
{
    int a, b;
    while(scanf("%d%d", &n, &m))
    {
        if(n==0 && m==0) break;
        memset(g, 0, sizeof(g));
        while(m--)
        {
            scanf("%d%d", &a, &b);
            g[a][b] = 1;
        }
        dfs_travel();
        for(int i = cnt-1; i > 0; i--)
            printf("%d ", topoSort[i]);
        printf("%d\n", topoSort[0]);
    }
    return 0;
}