PAT 甲级 1146 Topological Order

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原创最新推荐文章于 2026-06-15 09:15:41 发布 1.3k 阅读 AI 写同款· GEO 优化 ›

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1146 Topological Order （25 point(s)）

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

经验总结：

不知道为什么，PAT很喜欢从408里面找题目（可能是因为出题人也考过408吧0.0），题目不是很难，在输入边的时候，记录所有顶点的单向邻接点，并且记录所有点的入度。查询时，每输入一个点，就判断其入度是否为0，不为0，则不是拓扑排序，然后将此点指向的所有点的入度减一，查询至结束，输出即可~

AC代码

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=1010;
int n,m,k,indegree[maxn]={0},tdegree[maxn],a,b;
vector<int> adj[maxn];

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;++i)
	{
		scanf("%d%d",&a,&b);
		++indegree[b];
		adj[a].push_back(b);
	}
	scanf("%d",&k);
	bool f=false;
	for(int i=0;i<k;++i)
	{
		for(int i=1;i<=n;++i)
			tdegree[i]=indegree[i];
		bool flag=false;
		for(int j=0;j<n;++j)
		{
			scanf("%d",&a);
			if(tdegree[a]!=0)
				flag=true;
			for(int x=0;x<adj[a].size();++x)
				--tdegree[adj[a][x]];
		}
		if(flag==true)
			if(f==false)
			{
				printf("%d",i);
				f=true;
			}
			else
				printf(" %d",i);
	}
	return 0;
}

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