[SOJ] Ordering Tasks

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1940. Ordering TasksConstraints

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  

 拓撲排序

//AOV拓撲排序#include <iostream>#include <vector>#include <queue>#include <memory.h>using namespace std; int main() {  int numTestcases;  cin >> numTestcases;    while(numTestcases--)  {    int n, m;    cin >> n >> m;    int inDegree[n + 1];  //入度為0數組    int result[n];        //結果序列    vector<int> tasks[n + 1];   //每一組vector都有該結點的後繼結點    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<int, vector<int>, greater<int> > readyTasks;//使用最小優先隊列可以自動按照從小到大排序        for (int i = 1; i <= n; ++i)     {//先將所有根結點放進隊列中待選        if(inDegree[i] == 0)            readyTasks.push(i);    }    int numFinished = 0;        while(!readyTasks.empty())    {        int cur = readyTasks.top();        result[numFinished++] = cur;        readyTasks.pop();        vector<int>::iterator it;                for(it = tasks[cur].begin(); it != tasks[cur].end();it++)        {            int temp = *it;            inDegree[temp]--;            if(inDegree[temp] == 0)//當前趨結點全部完成時可以開始這個任務                readyTasks.push(temp);        }    }        for (int i = 0; i < n; ++i)     {        cout << result[i] << " ";    }    cout << endl;  }  return 0;}

　　

[SOJ] Ordering Tasks