Topological Order (判断是不是拓扑排序
所谓拓扑排序就是删除入度为0的点的顺序
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
#include<bits/stdc++.h>
#include <iostream>
#include <map>
#include <string>
#include<vector>
#include<stack>
using namespace std;
map<int ,vector<int> > mp;
int ru[100005];
int chu[100005];
int ru2[100005];
int chu2[100005];
int main() {
int n,m;
cin>>n>>m;
for(int i=0;i<m;i++)
{
int a,b;
cin>>a>>b;
chu[a]++;
ru[b]++;
mp[a].push_back(b);
}
int k;
cin>>k;
vector<int>ans;
for(int i=0;i<k;i++)
{
for(int u=1;u<=n;u++)
{
ru2[u]=ru[u];
chu2[u]=chu[u];
// cout<<ru2[u]<<endl;
}
int flag=1;
for(int j=0;j<n;j++)
{
int tmp;
cin>>tmp;
if(ru2[tmp]==0)
{
for(int t=0;t<mp[tmp].size();t++)
{
ru2[mp[tmp][t]]--;
}
}
else
{
// cout<<j<<endl;
flag=0;
}
}
if(flag==0) ans.push_back(i);
}
for(int i=0;i<ans.size()-1;i++)cout<<ans[i]<<" ";
cout<<ans[ans.size()-1];
return 0;
}