题目描述

Leetcode周赛6243. Minimum Fuel Cost to Report to the Capital

There is a tree (i.e., a connected, undirected graph with no cycles) structure country network consisting of n cities numbered from 0 to n - 1 and exactly n - 1 roads. The capital city is city 0. You are given a 2D integer array roads where roads[i] = [ai, bi] denotes that there exists a bidirectional road connecting cities ai and bi.

There is a meeting for the representatives of each city. The meeting is in the capital city.

There is a car in each city. You are given an integer seats that indicates the number of seats in each car.

A representative can use the car in their city to travel or change the car and ride with another representative. The cost of traveling between two cities is one liter of fuel.

Return the minimum number of liters of fuel to reach the capital city.

Example 1:

Input: roads = [[0,1],[0,2],[0,3]], seats = 5
Output: 3
Explanation: 
- Representative1 goes directly to the capital with 1 liter of fuel.
- Representative2 goes directly to the capital with 1 liter of fuel.
- Representative3 goes directly to the capital with 1 liter of fuel.
It costs 3 liters of fuel at minimum. 
It can be proven that 3 is the minimum number of liters of fuel needed.

Example 2:

Input: roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], seats = 2
Output: 7
Explanation: 

- Representative2 goes directly to city 3 with 1 liter of fuel.
- Representative2 and representative3 go together to city 1 with 1 liter of fuel.
- Representative2 and representative3 go together to the capital with 1 liter of fuel.
- Representative1 goes directly to the capital with 1 liter of fuel.
- Representative5 goes directly to the capital with 1 liter of fuel.
- Representative6 goes directly to city 4 with 1 liter of fuel.
- Representative4 and representative6 go together to the capital with 1 liter of fuel.
  It costs 7 liters of fuel at minimum. 
  It can be proven that 7 is the minimum number of liters of fuel needed.

思路

考虑每条边上至少需要多少辆车。

以 0 为根，设子树 x 的大小为 \(\textit{size}\)，那么它到它父节点这条边的「流量」是，那么就至少需要 \(\left\lceil\dfrac{\textit{size}}{\textit{seats}}\right\rceil\) 辆车。

求解向上取整的时候有个tick: ceil(x/a) = (x + a - 1)/a。

累加除了 \(x=0\)以外的值，就是答案。

这里需要注意的是，题目给出的是全部的边，我们应该根据边构建出图的临接表达，方便DFS搜索遍历。

代码

class Solution {
public:
    long long minimumFuelCost(vector<vector<int>>& roads, int seats) {
        //树总节点数比边多一
        int n = roads.size() + 1;
        vector<vector<int>> graph(n);
        for(vector<int>& road:roads){
            graph[road[0]].emplace_back(road[1]);
            graph[road[1]].emplace_back(road[0]);
        }
        long long res = 0;
        std::function<int(int,int)> dfs_lambda = [&](int node, int parent)->int {
            int size = 1;
            for(auto& child:graph[node]){
                if(child==parent) continue; //由于建立的是二维图，遍历时候不能往父节点遍历
                size += dfs_lambda(child,node);
            }
            if(node>0){
                res += (size + seats - 1)/seats; //向上取整
            }
            return size;
        };
        dfs_lambda(0,-1);
        return res;
    }
};

复杂度分析

• 时间复杂度：\(O(n)\)，其中 \(n\) 为 \(\textit{roads}\) 的长度。
• 空间复杂度：\(O(n)\)。