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[x] = q if person x has quietness q.

Now, return answer, where answer[x] = 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.

**分析：**

- 人的编号从0到n-1，每个人有一定的钱和度，并且都不相同
- 返回ans，ans[i]代表比i这个人钱多的人中，度最小的人的编号
- 数据（节点）长度500，边长度不超过250000，故显然对边分析时，n^2复杂度不可行

**思路：**

对于每个人可以直接用dfs找，但是也得用dp存一下。topsort也是一样，都得用dp。由于我对dfs还是运用不到位，这里先写下topsort方法吧。

对于example中的例子，关系如图：

可以看到2，4，5，6入度为0，即没有比他们更多钱的人，那么对于这种节点 ,ans[i] = i

对于3,则要从指向他的节点（4，5，6）以及他自身中找到quiet值最小的点index,ans[3] = min(3,4,5,6,key=lambda x: quiet[ans[x]]) ,依次类推即可

1 | import collections,Queue |