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

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题目大意：

代码：

#include<stdio.h>
#include<vector>
using namespace std;
int Map[1010][1010];
int dis[1010],dis1[1010];
vector<int> v;
int main()
{
    int i,j,n,m,k,t,x,y,l;
    scanf("%d %d",&n,&m);
    for(i=0;i<m;i++)
    {
        scanf("%d %d",&x,&y);
        Map[x][y]=1;
        dis1[y]++;
    }
    scanf("%d",&k);
    for(i=0;i<k;i++)
    {
        int flag=0;
        for(j=1;j<=n;j++)
        {
            dis[j]=dis1[j];
        }
        for(j=0;j<n;j++)
        {
            scanf("%d",&t);
            if(dis[t]!=0)
            {
                flag=1;
            }
            else
            {
                for(l=1;l<=n;l++)
                {
                    if(Map[t][l]==1)
                    {
                        dis[l]--;
                    }
                }
            }
        }
        if(flag==1)
        {
            v.push_back(i);
        }
    }
        for(i=0;i<v.size();i++)
        {
            if(i==0)
            {
                printf("%d",v[i]);
            }
            else
            {
                printf(" %d",v[i]);
            }
        }
    return 0;
}