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 <iostream>
#include <bits/stdc++.h>
using namespace std;
int g[1024][1024];//图
int du[1024];//入度
int now[1024];//已经完成任务的前驱的个数,当其与入度相等时,该结点可以执行。
int main()
{
memset(g, 0, sizeof(g));
memset(du, 0, sizeof(du));
int n, m;
cin>>n>>m;
for(int i=0; i<m; i++){
int a, b;
cin>>a>>b;
g[a][b] = 1;
du[b]++;
}
int k;
cin>>k;
int in = 0;
int re[128];
for(int u=0; u<k; u++){
memset(now, 0, sizeof(now));
int a[n+10];
for(int i=0; i<n; i++){
cin>>a[i];
}
for(int i=0; i<n; i++){
if(now[a[i]] == du[a[i]]){完成任务的前驱的个数与入度相等时,该结点可以执行。
for(int j=1; j<=n; j++){//注意这里是1~n,一开始写成了0~n-1......
if(g[a[i]][j] == 1){
now[j]++;//该结点所指向结点的完成任务的前驱的个数++
}
}
} else{
re[in++] = u;
break;
}
}
}
for(int i=0; i<in; i++){
printf(i!=in-1?"%d ":"%d\n",re[i]);
}
return 0;
}