1146 Topological Order 甲级 xp_xht123

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
6
5 2 3 6 4 1
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:

0 4 5

所谓拓扑顺序就是做事顺序，也就是做到下一个节点时必须经过上一个节点

解题思路：第一种方法，记录每个节点度数，根据给定的序列判断是否是拓扑顺序

第二种方法

首先，先将给定的集合按照给定顺序存入下标p
其次，遍历整张图，判断每个有向边的两个节点 a，b 的在p中的位置
最后，如果发现p[a] > p[b] 就不是拓扑排序
无需出度和入度的方法

第一种 

#include
#include
using namespace std;
 
const int N = 10100;
int n,m;
vector<int>gra[N];
vector<int>pan_cnt_front(N);//统计出现的次数
 
bool pan_is_toporder(vector<int>&exam)
{
    vector<int>pan_cnt_front_copy = pan_cnt_front;
    int len = exam.size();
    for(int i:exam)
    {
        if(pan_cnt_front_copy[i]) return false;
        for(int j:gra[i]) pan_cnt_front_copy[j]--;
    }
    return true;
}
 
int main()
{//拓扑顺序 做事顺序做到下一个节点时必须经过上一个节点
    scanf("%d %d",&n,&m);
    for(int i=0;i
    {
        int num1,num2;
        cin>>num1>>num2;
        gra[num1].push_back(num2);//有向图
        pan_cnt_front[num2]++;
    }
    int k;
    cin>>k;
    vector<int>res;
    for(int j=0;j
    {
        vector<int>exam(n);
        for(int i=0;i
            scanf("%d",&exam[i]);
        if(!pan_is_toporder(exam)) res.push_back(j);
    }
    for(int i=0;isize();i++)
    {
        if(i!=0) cout<<" ";
        cout<
    }
}

第二种

#include
#include
#include
 
using namespace std;
const int N = 1010 , M = 1e4 + 10;
 
int n , m;
int p[N];
struct edge
{
    int a , b;
}w[M];
 
int main()
{
    cin >> n >> m;
    for(int i = 0;i < m;i ++) cin >> w[i].a >> w[i].b;
    
    int k;
    cin >> k;
    vector<int>v;
    for(int i = 0;i < k;i ++)
    {
        for(int j = 1;j <= n;j ++)
        {
            int x;
            cin >> x;
            p[x] = j;
        }
        
        bool flag = true;
        for(int j = 0;j < m;j ++)
        {
            if(p[w[j].a] > p[w[j].b])
            {
                flag = false;
                break;
            }
        }
        if(!flag) v.push_back(i);
    }
    
    for(int i = 0;i < v.size();i ++)
    {
        if(i) cout << ' ';
        cout << v[i];
    }
    return 0;
}