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

<p>There is a meeting for the representatives of each city. The meeting is in the capital city.</p>

<p>There is a car in each city. You are given an integer <code>seats</code> that indicates the number of seats in each car.</p>

<p>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.</p>

<p>Return <em>the minimum number of liters of fuel to reach the capital city</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/22/a4c380025e3ff0c379525e96a7d63a3.png" style="width: 303px; height: 332px;" />
<pre>
<strong>Input:</strong> roads = [[0,1],[0,2],[0,3]], seats = 5
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
- Representative<sub>1</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>2</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>3</sub> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/11/16/2.png" style="width: 274px; height: 340px;" />
<pre>
<strong>Input:</strong> roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], seats = 2
<strong>Output:</strong> 7
<strong>Explanation:</strong> 
- Representative<sub>2</sub> goes directly to city 3 with 1 liter of fuel.
- Representative<sub>2</sub> and representative<sub>3</sub> go together to city 1 with 1 liter of fuel.
- Representative<sub>2</sub> and representative<sub>3</sub> go together to the capital with 1 liter of fuel.
- Representative<sub>1</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>5</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>6</sub> goes directly to city 4 with 1 liter of fuel.
- Representative<sub>4</sub> and representative<sub>6</sub> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/27/efcf7f7be6830b8763639cfd01b690a.png" style="width: 108px; height: 86px;" />
<pre>
<strong>Input:</strong> roads = [], seats = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> No representatives need to travel to the capital city.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>roads.length == n - 1</code></li>
	<li><code>roads[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>roads</code> represents a valid tree.</li>
	<li><code>1 &lt;= seats &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    //有點像車行發車的感覺,由乘客數量決定每個站點的發車數量,最後再由每站的發車數量來得知總體油量消耗,
    //return值表示乘客數量
    long long answer = 0;
    vector<vector<int>> adj;
    vector<bool> visited;
    long long DFS(int index, int seats) {
        visited[index] = true;
        //if(visited[index] == true)
        //    return 0;
        
        long long passengers = 1;
        visited[index] = true;
        for(int city : adj[index]) {
            if(visited[city] == false)
                passengers += DFS(city, seats);
        }
        //cout << "index: " << index << " passengers: " << passengers << endl;

        //回到capital就到終點了,可以不用再派車了!
        if(index != 0) {
            if(passengers % seats == 0)
                answer += passengers / seats;
            else
                answer += passengers / seats + 1;
        }
        return passengers;
    }

    long long minimumFuelCost(vector<vector<int>>& roads, int seats) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        //1.先建立相鄰矩陣(P.S.因為是雙向矩陣為了避免cycle,需要額外用visited array去阻止回溯)
        int nodes = roads.size() + 1; //題目條件有說邊數為點數-1,如果沒說的話就得用hash_map去存相鄰陣列
        visited.resize(nodes); 
        adj.resize(nodes);

        for(vector<int>& road : roads) {
            adj[road[0]].push_back(road[1]);
            adj[road[1]].push_back(road[0]);
        }

        //2.再透過DFS去累加答案
        DFS(0, seats);
        return answer;
    }
};
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