PAT Grade --A1146 TopologicalOrder [25]

This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topological order obtained from the given directed graph? Now you are supposed to write a program to test each of the options.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N ( $\leq 1,000), the number of vertices in the graph, and M (\leq 10,000), the number of directed edges. Then M lines follow, each gives the start and the end vertices of an edge. The vertices are numbered from 1 to N. After the graph, there is another positive integer K (\leq 100). Then K lines of query follow, each gives a permutation of all the vertices. All the numbers in a line are separated by a space.$

Output Specification:

Print in a line all the indices of queries which correspond to "NOT a topological order". The indices start from zero. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line. It is graranteed that there is at least one answer.

Sample Input:

6 8
1 2
1 3
5 2
5 4
2 3
2 6
3 4
6 4
5
1 5 2 3 6 4
5 1 2 6 3 4
5 1 2 3 6 4
5 2 1 6 3 4
1 2 3 4 5 6

Sample Output:

3 4

Solution: 
　　Use adjacency matrix has to preserve the matrix, and to calculate the degree of each node, determining the sequence of traversal, every elapse of a node determines that the node is not 0 degree, if not, the sequence topology description is not 
per after a node, which points to the node of -1, indicating that the parent node points to the node traversal is complete, thereby ensuring the entire sequence is a sequence of topological

 1 #include <iostream>
 2 #include <vector>
 3 #include <queue>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 
 8 int main()
 9 {    
10     int n, m, k;
11     cin >> n >> m;
12     vector<vector<int>>v(n + 1);
13     vector<int>in(n + 1, 0), temp, res;//节点的入度
14     for (int i = 0; i < m; ++i)
15     {
16         int a, b;
17         cin >> a >> b;
18         v[a].push_back(b);
19         in[b]++;
20     }
21     cin >> k;
22     for (int i = 0; i < k; ++i)
23     {
24         bool flag = true;
25         temp = in;        
26         for (int= J 0 ; J <n-; ++ J)
 27          {
 28              int X;
 29              CIN >> X;
 30              IF (TEMP [X] =! 0 ) = In Flag to false ;
 31 is              for (Auto A: V [X]) --temp [a]; // occur once a penetration Save 
32          }
 33 is          IF (! In Flag)
 34 is              res.push_back (I);
 35      }
 36      for ( int I = 0 ; I <res.size (); + + I)
 37 [          COUT << (I == 0 ? "" : " ") << res[i];    
38     return 0;
39 }