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

最后一个点会超时，此题代码仅供参考。

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<set>
#include<queue>
using namespace std;
int n, m, k;
int graph[1010][1010], tmp[1010][1010];
int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int v1, v2;
		scanf("%d%d", &v1, &v2);
		graph[v1][v2] = 1;
		tmp[v1][v2] = 1;
	}
	cin >> k;
	vector<int> ans;
	for (int i = 0; i < k; i++) {
		vector<int> qy(n);
		for (int j = 1; j <= n; j++) {
			for (int u = 1; u <= n; u++) {
				graph[j][u] = tmp[j][u];
			}
		}
		bool isout = 0;
		for (int j = 0; j < n; j++) {
			scanf("%d", &qy[j]);
			bool flag = 0;
			for (int u = 1; u <= n; u++) {
				if (graph[u][qy[j]] == 1) {
					flag = 1;
					if (isout == 0) {
						ans.push_back(i);
						isout = 1;
					}
					break;
				}
			}
			if (flag == 0) {
				for (int u = 1; u <= n; u++) {
					if (graph[qy[j]][u] == 1) graph[qy[j]][u] = 0;
				}
			}
		}
	}
	for (int i = 0; i < (int)ans.size(); i++) {
		if (i == 0) printf("%d", ans[i]);
		else printf(" %d", ans[i]);
	}
	return 0;
}