1146 Topological Order （25 分）

最新推荐文章于 2020-09-05 10:48:03 发布

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最新推荐文章于 2020-09-05 10:48:03 发布

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本文详细解析了拓扑排序算法的实现原理与应用，通过一个具体的编程实例，展示了如何判断给定序列是否为有效的拓扑排序。文章首先介绍了拓扑排序的基本概念，随后通过输入输出规格说明了算法的具体流程，包括读取图的顶点和边，计算各顶点的入度，并通过迭代删除入度为0的顶点来验证给定序列的有效性。

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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

只要理解拓扑的相关知识，这道题就简单了。

每次输出一个入度为0的点，然后将此点出发的路去除，终点的入度减一，然后再次遍历入度为0的点，全部的点全部按照这个规则输出，即为拓扑序列，如果最后依然有未输出的点，则不是拓扑序列。

#include<iostream>
#include<vector>
using namespace std;
int mp[1001][1001];
vector<int> v1(1001,0);
vector<int> v2(1001,0);
int n,m;
void judge(int a){
	for(int i = 1;i <= n;i++){
		if(mp[a][i] == 1){
			v2[i]--;
		}
	}
}
int main(){
	cin >> n >> m;
	for(int i = 1;i <= m;i++){
		int a,b;
		cin >> a >> b;
		mp[a][b] = 1;
		v1[b]++;
	}
	int k;
	vector<int> v;
	cin >> k;
	int book = 0;
	for(int i = 0;i < k;i++){
		int flag = 0;
		for(int j = 1;j <= n;j++){
			v2[j] = v1[j];
			//cout << " " << v2[j];
		}
		//cout << endl;
		for(int j = 1;j <= n;j++){
			int a;
			cin >> a;
			if(v2[a]==0){
			 	judge(a);
			}else{
				flag = 1;
				book++;
			}
		}
		if(flag){
			v.push_back(i);
		}
	}
	for(int i = 0;i < v.size();i++){
		if(i == 0){
			cout << v[i];
		}else{
			cout << " " << v[i];
		}
	}
	cout << endl;
	return 0;
}