851. Loud and Rich

Medium (Trung bình) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

There is a group of n people labeled from 0 to n - 1 where each person has a different amount of money and a different level of quietness.

You are given an array richer where richer[i] = [a_i, b_i] indicates that a_i has more money than b_i and an integer array quiet where quiet[i] is the quietness of the i^th person. All the given data in richer are logically correct (i.e., the data will not lead you to a situation where x is richer than y and y is richer than x at the same time).

Return an integer array 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 the 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 is not 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.

Example 2:

Input: richer = [], quiet = [0]
Output: [0]

Constraints:

n == quiet.length
1 <= n <= 500
0 <= quiet[i] < n
All the values of quiet are unique.
0 <= richer.length <= n * (n - 1) / 2
0 <= a_i, b_i < n
a_i!= b_i
All the pairs of richer are unique.
The observations in richer are all logically consistent.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Graph (Đồ thị)Bit Manipulation (Thao tác bit)

⏱️ Thời gian O(V+E)

💾 Không gian O(V)

💻 Lời Giải

C++ 0851-loud-and-rich.cpp

class Solution {
private:
    vector<vector<int>> adj;
    int n;
    vector<int> res;
    
public:
    int dfs(int u, vector<int> &quiet) {
        if (res[u] >= 0) {
            return res[u];
        }
        res[u] = u;
        for (int v : adj[u]) {
            if (quiet[res[u]] > quiet[dfs(v, quiet)]) {
                res[u] = res[v];
            }
        }
        return res[u];
    }
    vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
        this->n = quiet.size();
        adj.resize(n);
        
        for (auto r : richer) {
            int u = r[0], v = r[1];
            adj[v].push_back(u);
        }
        
        res.resize(n, -1);

        for (int i = 0; i < n; ++i) {
            res[i] = dfs(i, quiet);
        }
        
        return res;
    }
};