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
0 4 5
代码如下:
#includeusing namespace std; int main() { int n,m,k,v1,v2,a,flag=0,in[1010]; cin>>n>>m; vector v[1010]; for(int i=1;i<=m;i++) { cin>>v1>>v2; v[v1].push_back(v2); in[v2]++;//入度 } cin>>k; for(int i=0;i tin(in,in+n+1); for(int j=0;j >a; if(tin[a]!=0)judge=0;//入度不为0,不是拓扑序列,还不能输出 for(int it:v[a])tin[it]--;//每次选中某个点后要将它所指向的所有结点的入度-1 } if(judge==1)continue; printf("%s%d",flag==1?" ":"",i); flag=1; } cout< 运行结果如下:



