题目：
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
6
5 2 3 6 4 1
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:
0 4 5
题目大意：给一个有向图，判断给定序列是否是它的拓扑序列
思路：
用邻接表v存储这个有向图，并将每个节点的入度保存在in数组中。
对每一个要判断是否是拓扑序列的结点遍历，如果当前结点的入度不为0则表示不是拓扑序列（每次选中某个点后要将它所指向的所有结点的入度减去1），最后根据是否出现过入度不为0的点决定是否要输出当前的编号i
代码：（满分）
#include <iostream> #include <vector> using namespace std; int main() { int n, m, a, b, k, flag = 0, in[1010]; vector<int> v[1010]; scanf("%d %d", &n, &m); for (int i = 0; i < m; i++) { scanf("%d %d", &a, &b); v[a].push_back(b); in[b]++; } scanf("%d", &k); for (int i = 0; i < k; i++) { int judge = 1; vector<int> tin(in, in+n+1); for (int j = 0; j < n; j++) { scanf("%d", &a); if (tin[a] != 0) judge = 0; for (int it : v[a]) tin[it]--; } if (judge == 1) continue; printf("%s%d", flag == 1 ? " ": "", i); flag = 1; } return 0; }