Tideman算法自定义循环检测逻辑无法通过check50的原因咨询

// Lock pairs into the candidate graph in order, without creating cycles
void lock_pairs(void)
{
    // array[0] being true means that there is an edge or more pointing at candidate number 0
    bool array[candidate_count];
    for (int l = 0; l < pair_count; l++)
    {
        // add an edge, as long as it doesn't make a cycle
        if (!will_cycle(array, pairs[l].loser))
        {
            locked[pairs[l].winner][pairs[l].loser] = true;
            array[pairs[l].loser] = true;
        }
    }
}
// index is the index of the candidate the edge will be pointing at if it's added
// the function temporarily sets array[index] as true, like there's an edge pointing at it
// the function then iterates over all the array array, if they're all true it means there'll be a cycle.
// so it's undo the array[index] to false again.
// and if there's at least one false in the array array, it means there won't be a cycle
bool will_cycle(bool array[], int index) // return true if it will make a cycle
{
    array[index] = true;
    int flag = 0; // var for number of true candidates
    for (int i = 0; i < candidate_count; i++)
    {
        if (array[i])
            flag++;
    }

    if (flag == candidate_count) // they're all true
    {
        array[index] = false;
        return true;
    }
    return false;
}

我认为图中的循环意味着不存在源点（源点指没有入边的候选者），因此当array数组全为true时，所有候选者都有至少一条入边，即存在循环。

为何该逻辑无法通过check50中关于lock_pair跳过会导致循环的边的测试？我自测时该逻辑运行正常，为何必须使用DFS这类算法？

我测试了CS50官网pset页面背景中的Alice、Charlie和Bob的例子，跟踪后发现算法正确跳过了最后一条会导致循环的边，且能得出正确的获胜者。

内容的提问来源于stack exchange，提问作者Eyad Ayman