PAT甲级 1146 Topological Order 拓扑排序

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本文针对2018年研究生入学考试中的一道拓扑排序问题进行解析，提供了一种判断给定数列是否为特定有向图拓扑排序的算法实现。通过输入顶点数、边数及边的连接关系，结合一系列顶点排列，程序能够识别哪些排列不符合拓扑排序规则。

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1146 Topological Order

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.
在这里插入图片描述

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 guaranteed 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

题意为判断给定的数列是否为给定有向图的拓扑排序

根据拓扑排序的规则，拓扑序列中点n的前面必须出现图中射向点n的点

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <vector>

using namespace std;

#define SUM 1001

struct Point {//我不爱用vector数组
	vector<int> in_p;//射向该点的点
};

int n, m, k;
Point points[SUM];
vector<int> result;

int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int u, v;
		cin >> u >> v;
		points[v].in_p.push_back(u);
	}
	cin >> k;
	for (int i = 0; i < k; i++) {
		int book[SUM] = { 0 };//点是否出现过
		bool flag = false;
		for (int j = 1; j <= n; j++) {
			int p;
			cin >> p;
			book[p] = 1;
			for (int a : points[p].in_p) {
				if (book[a] == 0) {
					flag = true;
					break;
				}
			}
		}
		if (flag) {
			result.push_back(i);
		}
	}
	int len = result.size();
	for (int i = 1; i <= len; i++) {
		cout << result[i - 1];
		if (i != len) {
			cout << " ";
		}
	}
	cout << endl;
	return 0;
}