[Python](PAT)1146 Topological Order（25 分）

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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()

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