1146 Topological Order （25 分）判断拓扑序列

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
思路
　　定义一个count数组，保存每个结点对应的指向它的弧的个数，如上图中，count[1]=count[5]=0,count[2]=2,然后对
待检验序列进行遍历，对遍历到的每个结点：
　　　　　　　　　　　　　　　　　　　　如果当前结点的count数组值不等于0，则返回false
　　　　　　　　　　　　　　　　　　　　否则，对与该结点连接的每个结点的count值都减1

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<string.h>
#include<cstdio>
#include<cmath>
using namespace std;

vector<vector<int>> num;

bool check(int a[],int n,vector<int>& cnt)
{
    for(int i=0;i<n;i++)
    {
        int temp=a[i];
        if(cnt[temp]!=0)
            return false;
        for(auto &va:num[temp])
        {
            cnt[va]--;
           // cout<<va<<" "<<cnt[va]<<endl;
        }

    }
    return true;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    num.resize(n+1);
    vector<int> cnt(n+1,0);
   // cnt=(n+1,0);
    for(int i=0;i<m;i++)
    {
        int start,endL;
        cin>>start>>endL;
        num[start].push_back(endL);
        cnt[endL]++;
    }
    int k;
    cin>>k;
    int a[n];
    vector<int>result;
    vector<int>temp(n+1);
    for(int i=0;i<k;i++)
    {
        for(int j=0;j<n;j++)
        {
            temp[j+1]=cnt[j+1];
            cin>>a[j];
        }
        if(!check(a,n,temp))
            result.push_back(i);
    }
    cout<<result[0];
    for(int i=1;i<result.size();i++)
        cout<<" "<<result[i];
    return 0;
}