1146. Topological Order (25)

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
 
开门见山的题。
代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#define Max 1005
using namespace std;
int n,m,k,limit[Max],exa[Max],a,b,mp[Max][Max],ans[Max],ant;///limit存某个点前面有几个点先行
 
int check()
{
    int p[Max];
    for(int i = ;i <= n;i ++)///limit转到p
    {
        p[i] = limit[i];
    }
    for(int i = ;i < n;i ++)
    {
        if(p[exa[i]])///p为正表示顺序不合法
        {
            return ;
        }
        for(int j = ;j <= n;j ++)///如果合法 把受它限制的点 p都减1
        {
            if(mp[exa[i]][j])p[j] --;
        }
    }
    return ;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i = ;i < m;i ++)
    {
        scanf("%d%d",&a,&b);
        mp[a][b] = ;
        limit[b] ++;
    }
    scanf("%d",&k);
    for(int i = ;i < k;i ++)
    {
        for(int j = ;j < n;j ++)
        {
            scanf("%d",&exa[j]);
        }
        if(!check())
        {
            ans[ant ++] = i;
        }
    }
    for(int i = ;i < ant;i ++)
    {
        if(i)printf(" %d",ans[i]);
        else printf("%d",ans[i]);
    }
}