1146 Topological Order——PAT甲级

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
6
5 2 3 6 4 1
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:

0 4 5

 solution:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define endl '\n'
const int N=1e6+5;
int cnt=0,head[N],indegree[N];
vector<int>ans;
struct node
{
	int to,next;
}edge[N];
void addedge(int u,int v)
{
	cnt++;
	edge[cnt].to=v;
	edge[cnt].next=head[u];
	head[u]=cnt;
}
int n,m,k;
bool test(vector<int>&t)
{
	int testdegree[N];
	copy(indegree,indegree+N,testdegree);
	for(int i=0;i<t.size();i++)
	{
		if(testdegree[t[i]])return false;
		for(int u=head[t[i]];u>0;u=edge[u].next)
		{
			testdegree[edge[u].to]--;
		}
	}
	return true;
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int a,b;
		cin>>a>>b;
		addedge(a,b);
		indegree[b]++;
	}
	cin>>k;
	for(int i=0;i<k;i++)
	{
		vector<int>tmp(n);
		for(int i=0;i<n;i++)
		{
			cin>>tmp[i];
		}
		if(!test(tmp))ans.push_back(i);
	}
	for(int i=0;i<ans.size();i++)
	{
		if(i)cout<<' ';
		cout<<ans[i];
	}
	return 0;
}

         使用了链式前向星存储图的边，之后就是利用copy函数依次复制入度数组的数据，然后每经过一个结点就删除该结点的出度所对结点，根据题目给出数据元素依次删除，直至出现有入度不为0的元素或者删除完毕。