1146 topological order (25 points)

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 (≤ 1,000), the number of vertices in the graph, and M (≤ 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 (≤ 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

Give a picture , Then some sequences , Determine whether the sequence is the topological ordering of the whole graph
Judgment method ：

1. Record each node's read in at the time of input
2. When judging the sequence , Place a point in turn x Get out of here y Reduce your penetration by one , That is the x The point is y In front of the point , If it arrives y spot ,y The degree of penetration is still not 0, Explain that it should have been y Click the front point , It must be in y’ After the point .

#include <iostream>
#include <vector>
using namespace std;
const int n = 1e4 + 10;
int N, M, x, y, f;
vector<int> v[n];
int in[n];
int main()
{
    
    cin >> N >> M;
    while (M--)
    {
    
        cin >> x >> y;
        v[x].push_back(y);
        //  Record the degree of penetration 
        in[y]++;
    }
    cin >> M;
    for (int i = 0; i < M; i++)
    {
    
        //  Every time in Will change 
        vector<int> tin(in, in + N + 1);
        int judge = 0;
        for (int j = 0; j < N; j++)
        {
    
            cin >> x;
            if (tin[x])
                judge = 1;
            for (auto &e : v[x])
                tin[e]--;
        }
        if (judge == 0)
            continue;
        if (f++ != 0)
            cout << " ";
        cout << i;
    }
    return 0;
}