In a group of N people (labelled 0, 1, 2, ..., N-1), each person has different amounts of money, and different levels of quietness.

For convenience, we'll call the person with label x, simply "person x".

We'll say that richer[i] = [x, y] if person x definitely has more money than person y. Note that richer may only be a subset of valid observations.

Also, we'll say quiet = q if person x has quietness q.

Now, return answer, where answer = y if y is the least quiet person (that is, the person y with the smallest value of quiet[y]), among all people who definitely have equal to or more money than person x.

Example 1:

Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0] Output: [5,5,2,5,4,5,6,7] Explanation: answer[0] = 5. Person 5 has more money than 3, which has more money than 1, which has more money than 0. The only person who is quieter (has lower quiet[x]) is person 7, but it isn't clear if they have more money than person 0. answer[7] = 7. Among all people that definitely have equal to or more money than person 7 (which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x]) is person 7. The other answers can be filled out with similar reasoning. 

Note:

1. 1 <= quiet.length = N <= 500
2. 0 <= quiet[i] < N, all quiet[i] are different.
3. 0 <= richer.length <= N * (N-1) / 2
4. 0 <= richer[i][j] < N
5. richer[i][0] != richer[i][1]
6. richer[i]'s are all different.
7. The observations in richer are all logically consistent.

在一组 N 个人（编号为 0, 1, 2, ..., N-1）中，每个人都有不同数目的钱，以及不同程度的安静（quietness）。为了方便起见，我们将编号为 x 的人简称为 "person x "。如果能够肯定 person x 比 person y 更有钱的话，我们会说 richer[i] = [x, y] 。注意 richer 可能只是有效观察的一个子集。另外，如果 person x 的安静程度为 q ，我们会说 quiet[x] = q 。现在，返回答案 answer ，其中 answer[x] = y 的前提是，在所有拥有的钱不少于 person x 的人中，person y 是最安静的人（也就是安静值 quiet[y] 最小的人）。

提示：

• 1 <= quiet.length = N <= 500
• 0 <= quiet[i] < N，所有 quiet[i] 都不相同。
• 0 <= richer.length <= N * (N-1) / 2
• 0 <= richer[i][j] < N
• richer[i][0] != richer[i][1]
• richer[i] 都是不同的。
• 对 richer 的观察在逻辑上是一致的。

• 给出 2 个数组，richer 和 quiet，要求输出 answer，其中 answer = y 的前提是，在所有拥有的钱不少于 x 的人中，y 是最安静的人（也就是安静值 quiet[y] 最小的人）
• 由题意可知，richer 构成了一个有向无环图，首先使用字典建立图的关系，找到比当前下标编号富有的所有的人。然后使用广度优先层次遍历，不断的使用富有的人，但是安静值更小的人更新子节点即可。
• 这一题还可以用拓扑排序来解答。将 richer 中描述的关系看做边，如果 x > y，则 x 指向 y。将 quiet 看成权值。用一个数组记录答案，初始时 ans[i] = i。然后对原图做拓扑排序，对于每一条边，如果发现 quiet[ans[v]] > quiet[ans[u]]，则 ans[v] 的答案为 ans[u]。时间复杂度即为拓扑排序的时间复杂度为 O(m+n)。空间复杂度需要 O(m) 的数组建图，需要 O(n) 的数组记录入度以及存储队列，所以空间复杂度为 O(m+n)。

funcloudAndRich(richer [][]int, quiet []int) []int { edges:=make([][]int, len(quiet)) fori:=rangeedges { edges[i] = []int{} } indegrees:=make([]int, len(quiet)) for_, edge:=rangericher { n1, n2:=edge[0], edge[1] edges[n1] =append(edges[n1], n2) indegrees[n2]++ } res:=make([]int, len(quiet)) fori:=rangeres { res[i] =i } queue:= []int{} fori, v:=rangeindegrees { ifv==0 { queue=append(queue, i) } } forlen(queue) >0 { nexts:= []int{} for_, n1:=rangequeue { for_, n2:=rangeedges[n1] { indegrees[n2]--ifquiet[res[n2]] >quiet[res[n1]] { res[n2] =res[n1] } ifindegrees[n2] ==0 { nexts=append(nexts, n2) } } } queue=nexts } returnres }