PAT A1146 Topological Order （25 分）——拓扑排序，入度

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 <stdio.h>
#include <algorithm>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <queue>
using namespace std;
const int maxn=;
int a[maxn] ;
int n,m,k;
vector<int> v;
vector<int> adj[maxn];
int degree[maxn]={};
int store[maxn]={};
int main(){
    scanf("%d %d",&n,&m);
    for(int i=;i<=m;i++){
        int c1,c2;
        scanf("%d %d",&c1,&c2);
        degree[c2]++;
        adj[c1].push_back(c2);
    }
    scanf("%d",&k);
    for(int i=;i<k;i++){
        for(int j=;j<=n;j++){
            store[j]=degree[j];
        }
        int flag=;
        for(int j=;j<n;j++){
            int tmp;
            scanf("%d",&tmp);
            if(flag==){
                continue;
            }
            else{
                if(store[tmp]==){
                    for(int q=;q<adj[tmp].size();q++){
                        store[adj[tmp][q]]--;
                    }
                }
                else{
                    flag=;
                }
            }
        }
        if(flag==)v.push_back(i);
    }
    for(int i=;i<v.size();i++){
        printf("%d%s",v[i],i==v.size()-?"\n":" ");
    }
}

注意点：一开始没头绪要怎么做，翻了翻算法笔记，看了大佬的思路，发现原来这么方便，只要看进来的这个数入度是否为0。有一个坑就是不要一发现一个数不满足条件了就break，这样后面的输入就读不到了，要记录状态continue