PAT甲级1146 Topological Order (25分) 拓扑排序

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

方法1：
基本思路是拓补排序要求对于所有边u-v 则u在线性序列中出现在v之前
那么反过来判断是否满足拓扑排序，对于边u-v，在线性序列中u之前不能出现v
AC代码：

#include <iostream>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;


int main()
{
    //基本思路是拓补排序要求对于所有边u-v 则u在线性序列中出现在v之前
    //那么反过来判断是否满足拓扑排序，对于边u-v，在线性序列中u之前不能出现v
    
    int N,M;//顶点是1 -N；
    cin>>N>>M;
    int ver[N+1]= {0};
    vector<vector<int>> vec(N+1);//记录边
    for(int i=0; i<M; i++)
    {
        int u,v;
        cin>>u>>v;
        vec[u].push_back(v);//就是u前面不能有v
    }
    vector<int> outPut;//用来记录输出结果的

    int num;//需要判断的线性序列个数
    cin>>num;
    for(int q=0; q<num; q++)
    {
        int out=1;//先假设满足拓扑排序
        int sn[N];
        for(int i=0; i<N; i++)
        {
            cin>>sn[i];
        }
        
        int sign[N+1]= {0}; //记录之前出现的结点编号
        for(int i=0; i<N&&out==1; i++)
        {
            int checkNum=sn[i];//需要判断的编号
            if(i==0)
                sign[checkNum]=1;
            else
            {
                for(int p=0; p<vec[checkNum].size(); p++)
                //把编号checkNum 之前所有不能出现的编号遍历一遍
                {
                    //其实也就是对应边  checkNum->denyNum
                    int denyNum=vec[checkNum][p];
                    if(sign[denyNum]==1)//checkNum 之前出现了不能有的编号
                    {
                        out=0;
                        break;
                    }
                }
                if(out==1)
                {
                    sign[checkNum]=1;
                }
            }
        }
        if(out==0)
            outPut.push_back(q);
    }
    for(int i=0;i<outPut.size();i++){
        if(i==0)
            cout<<outPut[i];
        else
            cout<<' '<<outPut[i];
    }


    return 0;
}

方法二： 参考了柳神 https://blog.csdn.net/liuchuo/article/details/79805295 思路
我们可以从入度的角度考虑，因为这里是所有结点的线性序列，所以每个结点之前的所有结点对应边的末端结点 入度-1 后，所有结点的入度为0
比如B选项2之前 有 5 和1 结点，他们对应边的末端结点就是2 3 4 ，那么就把 2 3 4结点的入度-1，你会发现2结点的入度刚好为0了，而D结点的2结点入度为1 那么就不满足拓扑排序。

代码可以参考柳神的，状态不好，不想撸了。。。。