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
题目大意：拓扑排序，给定一个有向图，判断哪个序列是拓扑序列，把给定的序列从0 - n-1进行编号，输出不是拓扑排序的编号。
这道题用邻接表来存储有向图会比较方便，因为涉及到入度的计算。建立两个vector，一个存放邻接表，另一个存放入度。之后给定的序列中，每读入一个数字，判断其入度是否为0，若不是，则不是拓扑排序，如果是，将其出边的所有点的入度-1，不断循环即可。
代码：
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int main(){
int pcnt, vcnt;
cin >> pcnt >> vcnt;
vector<int> inv(pcnt + 1, 0);
vector<vector<int> > G(pcnt + 1);
for(int i = 0; i < vcnt; i++){
int l, r;
cin >> l >> r;
inv[r]++;
G[l].push_back(r);
}
int qcnt;
cin >> qcnt;
vector<int> ans;
for(int i = 0; i < qcnt; i++){
vector<int> tmpinv = inv;
int flag = false;
for(int i = 0; i < pcnt; i++){
int n;
cin >> n;
if(tmpinv[n] != 0) flag = true;
if(flag == false)
for(int it : G[n]) tmpinv[it]--;
}
if(flag) ans.push_back(i);
}
for(int i = 0; i < ans.size(); i++){
if(i) cout << ' ';
cout << ans[i];
}
cout << endl;
return 0;
}