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题目大意
给定一个有向图，再给定若干个查询，每个查询中包括一个有向拓扑序列。如果这个查询中的有向拓扑序列不是这个有向图的有向拓扑序列，则输出该查询的编号。
分析
关于邮箱拓扑图的介绍，可以看这里有向拓扑图
使用start字典来记录每个节点的向下连接点。
使用to字典记录每个节点的入度。
遍历每一个查询，遍历查询中的序列中的每一个元素。
如果当前元素的入度不为0，则说明这个点还有前驱点，不应该弹出。所以把这个查询的编号压入result数组。
否则的话，把这个应该弹出的节点对应的各个向下连接点的入度减一。
最终输出所有的非有向拓扑图查询编号即可
Python实现
def main():
line = input().split(" ")
n,m = int(line[0]), int(line[1])
start = {key : [] for key in range(n)}
to = {key : 0 for key in range(n)}
for x in range(m):
line = input().split(" ")
a, b = int(line[0])-1, int(line[1])-1
start[a].append(b)
to[b] += 1
k = int(input())
result = []
for x in range(k):
line = input().split(" ")
num = [int(x)-1 for x in line]
t = start.copy()
h = to.copy()
for i in num:
if h[i] !=0:
result.append(x)
break
else:
for j in t[i]:
h[j] -= 1
print(*result)
if __name__ == "__main__":
main()