851. Loud and Rich #

题目 #

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

题目大意 #

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。

解题思路 #

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)。

代码 #

func loudAndRich(richer [][]int, quiet []int) []int {
	edges := make([][]int, len(quiet))
	for i := range edges {
		edges[i] = []int{}
	}
	indegrees := make([]int, len(quiet))
	for _, edge := range richer {
		n1, n2 := edge[0], edge[1]
		edges[n1] = append(edges[n1], n2)
		indegrees[n2]++
	}
	res := make([]int, len(quiet))
	for i := range res {
		res[i] = i
	}
	queue := []int{}
	for i, v := range indegrees {
		if v == 0 {
			queue = append(queue, i)
		}
	}
	for len(queue) > 0 {
		nexts := []int{}
		for _, n1 := range queue {
			for _, n2 := range edges[n1] {
				indegrees[n2]--
				if quiet[res[n2]] > quiet[res[n1]] {
					res[n2] = res[n1]
				}
				if indegrees[n2] == 0 {
					nexts = append(nexts, n2)
				}
			}
		}
		queue = nexts
	}
	return res
}

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