1146 Topological Order （25 分）

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

C++:

/*
 @Date    : 2018-08-08 20:54:44
 @Author  : 酸饺子 ([email protected])
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://pintia.cn/problem-sets/994805342720868352/problems/994805343043829760
 */

#include <iostream>
#include <cstdio>
#include <vector>

using namespace std;

const int MAXN = 1005;
int N, M;
bool G[MAXN][MAXN];
vector<int> indegree(MAXN);

int main(int argc, char const *argv[])
{
    for (int i = 0; i < MAXN; ++i)
    {
        for (int j = 0; j < MAXN; ++j)
        {
            G[i][j] = false;
        }
        indegree[i] = 0;
    }

    scanf("%d %d", &N, &M);
    for (int i = 0; i < M; ++i)
    {
        int v, w;
        scanf("%d %d", &v, &w);
        G[v][w] = true;
        ++indegree[w];
    }

    int K;
    vector<int> ret;
    scanf("%d", &K);
    for (int i = 0; i < K; ++i)
    {
        bool top = true;
        vector<int> tempIndegree(indegree);
        for (int j = 0; j < N; ++j)
        {
            int v;
            scanf("%d", &v);
            if (top)
            {
                if (tempIndegree[v] == 0)
                {
                    for (int k = 1; k <= N; ++k)
                    {
                        if (G[v][k])
                        {
                            --tempIndegree[k];
                        }
                    }
                }
                else
                    top = false;
            }
        }
        if (!top)
        {
            ret.push_back(i);
        }
    }

    int output = 0;
    for (auto &i : ret)
    {
        if (output++)
        {
            putchar(' ');
        }
        printf("%d", i);
    }
    putchar('\n');
    return 0;
}