有向图的拓扑排序问题

【问题描述】给定有向图的adjacent list，写出其拓扑排序结果
【输入形式】节点数目n，和有向图的 adjacent list。每个节点i的adjacent list 表示为数组[i,x,x]; 如第0个顶点和第1,5个有有向边，表示为[0 1 5]

【输出形式】拓扑排序结果

【样例输入】

10

[0 6 1 4]

[1 2]

[2 7 5]

[3 8]

[4 5]

[5 9]

[6 3 2]

[7 8]

[8 9]

[9 ]

【样例输出】[ 0 6 1 4 3 2 7 5 8 9]
我的代码：

img

GraphList(int numVert)
{
    numVertice = numVert;
    int i;
    graph = new list * [numVert];
    for (i = 0; i < numVert; i++)
        graph[i] = new list;

    char ch = '[';
    for (i = 0; i < numVert; i++)
    {
        cin.get(ch);
        cin >> graph[i]->n;
        graph[i]->next = NULL;
        list* p = new list();
        p = graph[i];
        while (ch != ']')
        {
            list* q = new list();
            cin.get(ch);

            cin >> q->n;
            q->next = NULL;
            p->next = q;
            p = p->next;
        }
    }
}

int first(int v)
{
    if (graph[v]->next)return graph[v]->next->n;
    return -1;
}
int next(int v1, int v2)
{
    if (v1 == v2)return graph[v1]->next->n;
    if (graph[v2]->next)return graph[v2]->next->n;
    return -1;
}

void topsort(queue<int>* Q)
{
    int* indegree = new int[numVertice];
    int v, w;

    for (v = 0; v < numVertice; v++)
        indegree[v] = 0;
    for (v = 0; v < numVertice; v++)
        for (w = this->first(v); w != -1; w = this->next(v, w))
            indegree[w]++;
    for (v = 0; v < numVertice; v++)
        if (indegree[v] == 0)
            Q->push(v);

    cout << '[';
    while (Q->size() != 0)
    {
        v = Q->front();
        cout << " "<< v ;
        Q->pop();

        for (w = this->first(v); w != -1; w = this->next(v, w))
        {
            indegree[w]--;
            if (indegree[w] == 0)
                Q->push(w);
        }
    }
    cout << ']';

}

};

int main()
{
int n; cin >> n;
GraphList* graph = new GraphList(n);
queue* q = new queue;
graph->topsort(q);
}
代码问题出在构造函数GraphList中，具体是：
cin.get(ch)得到的ch是一个换行符（要连续三次cin.get(ch) ch 才为'[' )，后面的while循环也出问题。

烦请各位指出我的问题。