1940. Ordering Tasks

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

 

 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

 

There are multiple test cases. The first line contains an integer T, indicating the number of test cases. Each test case begins with a line containing two integers, 1 <= n <= 100000 and 1 <= m <= 100000. 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. It is guaranteed that no task needs to be executed before itself either directly or indirectly. 

 

Output

 

For each test case, print a line with n integers representing the tasks in a possible order of execution. To separate them, print exactly one space after each integer. If there are multiple solutions, output the smallest one by lexical order.  

 

Sample Input

15 53 44 13 22 45 3

Sample Output

5 3 2 4 1

 

拓扑排序

//AOV拓扑排序#include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std; int main() {  int numTestcases;  cin >> numTestcases;    while(numTestcases--)  {    int n, m;    cin >> n >> m;    int inDegree[n + 1];  //入度为0数组    int result[n];        //结果序列    vector
          
            tasks[n + 1];   //每一组vector都有该结点的后继结点    memset(inDegree, 0, sizeof(inDegree));  //初始化        for (int i = 0; i < m; ++i)    {        int a, b;        cin >> a >> b;        inDegree[b]++;        tasks[a].push_back(b);    }        priority_queue
           

, greater

> readyTasks;//使用最小优先队列可以自动按照从小到大排序 for (int i = 1; i <= n; ++i) {//先将所有根结点放进队列中待选 if(inDegree[i] == 0) readyTasks.push(i); } int numFinished = 0; while(!readyTasks.empty()) { int cur = readyTasks.top(); result[numFinished++] = cur; readyTasks.pop(); vector

::iterator it; for(it = tasks[cur].begin(); it != tasks[cur].end();it++) { int temp = *it; inDegree[temp]--; if(inDegree[temp] == 0)//当前趋结点全部完成时可以开始这个任务 readyTasks.push(temp); } } for (int i = 0; i < n; ++i) { cout << result[i] << " "; } cout << endl; } return 0;}