欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页

PAT (Advanced Level) Practice 1146 Topological Order (25分) (拓扑序列的判断)

程序员文章站 2022-07-08 22:20:18
...

1.题目

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.

PAT (Advanced Level) Practice 1146 Topological Order (25分) (拓扑序列的判断)

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

2.题目分析

1.拓扑序列的判断:拓扑序列,即有向无环,每次找一个入度为零的点,将所有和它相连的点的入度减一(删除相连的边),重复此步骤(有n个点,每次选择一个入度为0的点,进行n次)。如果这时还有入度不为零的点,证明有环。

2.注意要是进行多次判断的话,记得不能动原来的in数组,要另外设置备份数组进行修改判断

3.代码

#include<iostream>
#include<vector>
#include <algorithm>
using namespace std;
#define MAX 1050
int n, m,k;
int edges[MAX][MAX];
int in[MAX];
int main()
{
	scanf("%d %d", &n, &m);
	int a, b;
	for (int i = 1; i <= m; i++)
	{
		scanf("%d %d", &a, &b);
		edges[a][b] = 1;
		in[b]++;
	}
	scanf("%d", &k);
	int space = 0;
	for (int i = 1; i <= k; i++)
	{
        bool ok=true;
		vector<int>temp(in, in + n + 1);
		for (int j = 1; j <= n; j++)
		{
			int c;
			scanf("%d", &c);
			if (temp[c] != 0)
			ok=false;
			for (int s = 1; s <= n; s++)
				if (edges[c][s] != 0)
                temp[s]--;
		}
            if(!ok)
            {
				if (space == 0) { printf("%d", i - 1); space++; }
				else printf(" %d", i - 1);
			}
	}
}