1146 Topological Order (25 分) 判断是否为拓扑序列

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.

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

听说有一题考拓扑排序的，特地来看看，果然很简单。直接考拓扑排序太老套，本题懂拓扑排序的大致算法即可做

#include<bits/stdc++.h>
using namespace std;

const int maxn=1010;
vector<int> Adj[maxn];
int n,m;
int inDegree[maxn]={0},A[maxn];

void print(int a[]){
  for(int i=1;i<=n;i++) cout<<a[i]<<" ";
  cout<<endl;
}

int main(){
//  freopen("in.txt","r",stdin);
  cin>>n>>m;
  int a,b;
  while(m--){
    scanf("%d%d",&a,&b);
    Adj[a].push_back(b);//a->b
    inDegree[b]++;
  } 
  int k;
  vector<int> ans;
  int degree[maxn];
  
  cin>>k;
  for(int t=0;t<k;t++){
    for(int i=0;i<n;i++){
      scanf("%d",&A[i]);
    }
    for(int i=1;i<=n;i++) degree[i]=inDegree[i];
    
    for(int i=0;i<n;i++){
      if(degree[A[i]]!=0){
        ans.push_back(t);
        break;
      }
      vector<int> ve=Adj[A[i]];
      for(int j=0;j<ve.size();j++){
        degree[ve[j]]--;
      }
    }
  }
  
  for(int i=0;i<ans.size();i++){
     if(i>0) printf(" ");
     printf("%d",ans[i]); 
  } 
  
  return 0;
}