如果图中存在有向环，则不存在拓扑排序，反之则存在。不包含有向环的有向图成为有向无环图。可借助DFS完成拓扑排序：在访问完一个结点之后把它加到当前拓扑排序的首部。

int c[maxn];
int topo[maxn],t;
bool dfs(int u){
    c[u]=-1;//访问标志
    for(int v=0;v<n;v++){
        if(G[u][v]){
            if(c[v]<0) return false;//存在有向环，失败退出
            else if(!c[v]&&!dfs(v)) return false;
        }
    }
    c[u]=1;
    topo[--t]=u;
    return true;
}
bool toposort(){
    t=n;
    memset(c,0,sizeof(c));
    for(int u=0;u<n;u++){
        if(!c[u]){
            if(!dfs(u)) return false;
        }
    }
    return true;
}

c[u]=0表示从来没有访问过（从来没有调用过dfs(u)）;c[u]=1表示已经访问过，并且还递归访问过它的所有子孙（即dfs(u)曾被调用过，并已返回）；c[u]=-1表示正在访问（即递归调用dfs(u)正在帧栈中，尚未返回）。

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//不是拓扑排序的序号

题目大意：给一个有向图，判断给定序列列是否是拓扑序列～
分析：用邻接表v存储这个有向图，并将每个节点的入度保存在in_degree数组中。对每一个要判断是否是拓扑序列的结点遍历，如果当前结点的入度不为0则表示不是拓扑序列，每次选中某个点后要将它所指向的所有结点的入度-1，最后根据是否出现过入度不为0的点决定是否要输出当前的编号i～flag是用来判断之前是否输出过现在是否要输出空格的～judge是用来判断是否是拓扑序列的～

#include <iostream>
#include <vector>

using namespace std;
const int N = 1010;
int main(){
    //freopen("1.txt","r",stdin);
    int n,m,x,y,k;
    int in_degree[N]={0},flag=0;
    scanf("%d%d",&n,&m);
    vector<int>v[n+1];
    while(m--){
        scanf("%d%d",&x,&y);
        v[x].push_back(y);
        in_degree[y]++;
    }
    scanf("%d",&k);
    for(int i=0;i<k;i++){
        int judge=1;
        vector<int>back_up(in_degree,in_degree+n+1);
        for(int j=0;j<n;j++){
            scanf("%d",&x);
            if(back_up[x]!=0) judge=0;
            for(auto y:v[x]) back_up[y]--;
        }
        if(judge==1) continue;
        printf("%s%d",flag==1?" ":"",i);
        flag=1;
    }
    return 0;
}