1146 Topological Order (25 minutes) (determines whether the ordered sequence of topology)

 

1146 Topological Order (25分)

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

Subject to the effect: given a set of graphs directed edge, and then give k genome sequence, which allows you to not judge topological sorting 

Thinking: with FIG storage array vector, inDegree [] array of the recording of each dot, arr [] array stores the sequence to be checked

Arr each loop check point, if it is not 0-degree BREAK, all the points of each point of the point minus 1

Complete code:

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
vector<int> g[1001];
int n,m,k,arr[1001],inDegree[1001];
int main(){
    cin>>n>>m;
    int u,v,temp[1001] = {0},res[101];
    int cnt = 0;
    for(int i = 0;i<m;i++){
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        temp[v]++;
    }
    cin>>k;
    for(int i = 0;i<k;i++){
        //记得每次都要重新赋值inDegree数组
        for(int j = 1;j<=n;j++){
            inDegree[j] = temp[j];
        }
        for(int j = 1;j<=n;j++){
            scanf("%d",&arr[j]);
        }
        for(int j = 1;j<=n;j++){
            //入度不为0就不能形成拓扑排序
            if(inDegree[arr[j]]!=0){
                res[cnt++] = i;
                break;
            }
            //该点指向的所有点入度减1
            for(int u = 0;u<g[arr[j]].size();u++){
                int v = g[arr[j]][u];
                inDegree[v]--;
            }
        }
    }
    bool flag = false;
    for(int i = 0;i<cnt;i++){
        if(flag==false){
            cout<<res[i];
            flag = true;
        }
        else {
            cout<<" "<<res[i];
        }
    }
    return 0;
}