PAT 1146 Topological Order （25 分）

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本文介绍了一个竞赛题目，要求根据给定的有向图判断一系列顶点序列是否构成正确的拓扑排序。文章提供了算法思路，采用邻接表存储图结构，并通过检查每个顶点的入度来验证序列的正确性。

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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

题意：根据给定的有向图判断给定的拓扑排序序列是否正确

解题思想：以邻接表的形式存储有向图，用数组存储每一个点的入度，遍历给点拓扑排序序列中的结点，如果其入度为0，则是拓扑排序，否则不是，遍历的结点选定后要将其所指向的结点的入度减一。在遍历的过程中若出现入度不为零的点则可判定此非拓扑排序，可输出其序列编号；同时输出之前还要判断之前是否有过输出，若有则要加以空格相隔。

#include <iostream>
#include <vector>
#include <stdio.h>
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]--;
        }