PAT甲级——A1146 TopologicalOrder【25】

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

Solution：
　　使用邻接矩阵来保存这个有向矩阵，并且把每个节点的入度计算，遍历判断的序列，每经过一个节点就判断该节点入度是不是为0，若不是，说明不是拓扑序列
每经过一个节点，将其指向节点的入度-1，表明指向节点的父节点遍历完毕，从而保证了整个序列是个拓扑序列

 #include <iostream>
 #include <vector>
 #include <queue>
 #include <algorithm>
 using namespace std;

 int main()
 {
     int n, m, k;
     cin >> n >> m;
     vector<vector<int>>v(n + );
     vector<int>in(n + , ), temp, res;//节点的入度
     for (int i = ; i < m; ++i)
     {
         int a, b;
         cin >> a >> b;
         v[a].push_back(b);
         in[b]++;
     }
     cin >> k;
     for (int i = ; i < k; ++i)
     {
         bool flag = true;
         temp = in;
         for (int j = ; j < n; ++j)
         {
             int x;
             cin >> x;
             if (temp[x] != )flag = false;
             for (auto a : v[x])--temp[a];//出现一次入度减一
         }
         if (!flag)
             res.push_back(i);
     }
     for (int i = ; i < res.size(); ++i)
         cout << (i ==  ? "" : " ") << res[i];
     return ;
 }