PAT 1146 Topological Order （25 分）

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本文解析了2018年研究生入学考试中的一道拓扑排序题目，详细介绍了如何通过编程判断给定的顶点排列是否为有效的拓扑排序。输入包括图的顶点数、边数及边的连接，以及多个顶点排列，输出不符合拓扑排序条件的排列索引。

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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.
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 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

解析

#include<iostream>
#include<string>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
vector<vector<int>> G;
vector<int> InDegree;
bool Topological(const vector<int>& top) {
	vector<int> In(InDegree.cbegin(), InDegree.cend());
	set<int> Q;
	for (int i = 1; i < In.size(); i++) {
		if (In[i] == 0)
			Q.insert(i);
	}
	auto it = top.begin();
	while (!Q.empty()) {
		if (Q.find(*it) != Q.cend()) {
			Q.erase(*it);
			for (auto x : G[*it]) {
				In[x]--;
				if (In[x] == 0)
					Q.insert(x);
			}
		}
		else
			return false;
		it++;
	}
	return it == top.cend();
}
int main()
{
	int N, M;
	scanf("%d %d", &N, &M);
	G.resize(N+1);
	InDegree.resize(N+1, 0);
	int v1, v2;
	for (int i = 0; i < M; i++) {
		scanf("%d %d", &v1, &v2);
		G[v1].push_back(v2);
		InDegree[v2]++;
	}
	int K;
	scanf("%d", &K);
	vector<int> top(N);
	vector<int> NOT;
	for (int i = 0; i < K; i++) {
		for (int i = 0; i < N; i++)
			scanf("%d", &top[i]);
		if (Topological(top) == false)
			NOT.push_back(i);
	}
	for (int i = 0; i < NOT.size(); i++) {
		printf("%d%c", NOT[i], i + 1 == NOT.size() ? '\n': ' ');
	}
}