lc课程表2 拓扑排序的理解和反三

const int N=4010;
class Solution {
public:
int h[N], e[N], ne[N], idx;
int q[N];              
int d[N];

void add(int a, int b)
{
    e[idx]=b, ne[idx]=h[a], h[a]=idx++;
}

bool topsort(int n)
{
    int hh=0,tt=-1;

    for (int i = 0; i < n; ++i) {
        if(!d[i])
            q[++tt]=i;
    }

    while(hh<=tt)
    {

        int t=q[hh++];

        for(int i=h[t];i!=-1;i=ne[i])  
        {
            int j=e[i];     

            if(--d[j]==0)         
                q[++tt]=j;
        }
    }

    return tt==n-1;
}
    vector<int> findOrder(int num, vector<vector<int>>& p) {
        vector<int> ans(num);
        memset(h, -1, sizeof h);
        for(auto x:p)
        {
        add(x[1],x[0]);
        d[x[0]]++;
        }
        if(topsort(num))
        {
            for (int i = 0; i < num; ++i) 
            ans[i]=q[i];
            return ans;
        }
    return {};
    }
};