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                # 1168. 水資源分配優化 ## 題目地址(1168. 水資源分配優化) <https://leetcode.com/problems/optimize-water-distribution-in-a-village/> ## 題目描述 ``` <pre class="calibre18">``` 村莊內有n戶人家,我們可以通過挖井或者建造水管向每家供水。 對于每戶人家i,我們可以通過花費 wells[i] 直接在其房內挖水井,或者通過水管連接到其他的水井。每兩戶住戶間鋪設水管的費用通過 pipes 數組表示。 pipes[i] = [house1, house2, cost] 表示住戶1到住戶2間鋪設水管的費用為cost。 請求出所有住戶都能通水的最小花費。 示例1: 輸入: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]] 輸出: 3 解釋: The image shows the costs of connecting houses using pipes. The best strategy is to build a well in the first house with cost 1 and connect the other houses to it with cost 2 so the total cost is 3. 提示: 1 <= n <= 10000 wells.length == n 0 <= wells[i] <= 10^5 1 <= pipes.length <= 10000 1 <= pipes[i][0], pipes[i][1] <= n 0 <= pipes[i][2] <= 10^5 pipes[i][0] != pipes[i][1] ``` ``` ## 前置知識 - 圖 - 最小生成樹 ## 公司 - 暫無 ## 思路 ![](https://img.kancloud.cn/84/6f/846f99c2d04bd22b2bfa3c522f7ed34b_450x408.jpg) 題意,在每個城市打井需要一定的花費,也可以用其他城市的井水,城市之間建立連接管道需要一定的花費,怎么樣安排可以花費最少的前灌溉所有城市。 這是一道連通所有點的最短路徑/最小生成樹問題,把城市看成圖中的點,管道連接城市看成是連接兩個點之間的邊。這里打井的花費是直接在點上,而且并不是所有 點之間都有邊連接,為了方便,我們可以假想一個點`(root)0`,這里自身點的花費可以與 `0` 連接,花費可以是 `0-i` 之間的花費。這樣我們就可以構建一個連通圖包含所有的點和邊。 那在一個連通圖中求最短路徑/最小生成樹的問題. 參考延伸閱讀中,維基百科針對這類題給出的幾種解法。 解題步驟: 1. 創建 `POJO EdgeCost(node1, node2, cost) - 節點1 和 節點2 連接邊的花費`。 2. 假想一個`root` 點 `0`,構建圖 3. 連通所有節點和 `0`,`[0,i] - i 是節點 [1,n]`,`0-1` 是節點 `0` 和 `1` 的邊,邊的值是節點 `i` 上打井的花費 `wells[i]`; 4. 把打井花費和城市連接點轉換成圖的節點和邊。 5. 對圖的邊的值排序(從小到大) 6. 遍歷圖的邊,判斷兩個節點有沒有連通 (`Union-Find`), - 已連通就跳過,繼續訪問下一條邊 - 沒有連通,記錄花費,連通節點 7. 若所有節點已連通,求得的最小路徑即為最小花費,返回 8. 對于每次`union`, 節點數 `n-1`, 如果 `n==0` 說明所有節點都已連通,可以提前退出,不需要繼續訪問剩余的邊。 > 這里用加權Union-Find 判斷兩個節點是否連通,和連通未連通的節點。 舉例:`n = 5, wells=[1,2,2,3,2], pipes=[[1,2,1],[2,3,1],[4,5,7]]` 如圖: ![](https://img.kancloud.cn/72/44/72449e866d946ddadd3f068e83f716fb_1440x1080.jpg) 從圖中可以看到,最后所有的節點都是連通的。 **復雜度分析** - 時間復雜度: `O(ElogE) - E 是圖的邊的個數` - 空間復雜度: `O(E)` > 一個圖最多有 `n(n-1)/2 - n 是圖中節點個數` 條邊 (完全連通圖) ## 關鍵點分析 1. 構建圖,得出所有邊 2. 對所有邊排序 3. 遍歷所有的邊(從小到大) 4. 對于每條邊,檢查是否已經連通,若沒有連通,加上邊上的值,連通兩個節點。若已連通,跳過。 ## 代碼 (`Java/Python3`) *Java code* ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">OptimizeWaterDistribution</span> </span>{ <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-keyword">int</span> <span class="hljs-title">minCostToSupplyWater</span><span class="hljs-params">(<span class="hljs-keyword">int</span> n, <span class="hljs-keyword">int</span>[] wells, <span class="hljs-keyword">int</span>[][] pipes)</span> </span>{ List<EdgeCost> costs = <span class="hljs-keyword">new</span> ArrayList<>(); <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-params">1</span>; i <= n; i++) { costs.add(<span class="hljs-keyword">new</span> EdgeCost(<span class="hljs-params">0</span>, i, wells[i - <span class="hljs-params">1</span>])); } <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span>[] p : pipes) { costs.add(<span class="hljs-keyword">new</span> EdgeCost(p[<span class="hljs-params">0</span>], p[<span class="hljs-params">1</span>], p[<span class="hljs-params">2</span>])); } Collections.sort(costs); <span class="hljs-keyword">int</span> minCosts = <span class="hljs-params">0</span>; UnionFind uf = <span class="hljs-keyword">new</span> UnionFind(n); <span class="hljs-keyword">for</span> (EdgeCost edge : costs) { <span class="hljs-keyword">int</span> rootX = uf.find(edge.node1); <span class="hljs-keyword">int</span> rootY = uf.find(edge.node2); <span class="hljs-keyword">if</span> (rootX == rootY) <span class="hljs-keyword">continue</span>; minCosts += edge.cost; uf.union(edge.node1, edge.node2); <span class="hljs-title">// for each union, we connnect one node</span> n--; <span class="hljs-title">// if all nodes already connected, terminate early</span> <span class="hljs-keyword">if</span> (n == <span class="hljs-params">0</span>) { <span class="hljs-keyword">return</span> minCosts; } } <span class="hljs-keyword">return</span> minCosts; } <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">EdgeCost</span> <span class="hljs-keyword">implements</span> <span class="hljs-title">Comparable</span><<span class="hljs-title">EdgeCost</span>> </span>{ <span class="hljs-keyword">int</span> node1; <span class="hljs-keyword">int</span> node2; <span class="hljs-keyword">int</span> cost; <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-title">EdgeCost</span><span class="hljs-params">(<span class="hljs-keyword">int</span> node1, <span class="hljs-keyword">int</span> node2, <span class="hljs-keyword">int</span> cost)</span> </span>{ <span class="hljs-keyword">this</span>.node1 = node1; <span class="hljs-keyword">this</span>.node2 = node2; <span class="hljs-keyword">this</span>.cost = cost; } <span class="hljs-params">@Override</span> <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-keyword">int</span> <span class="hljs-title">compareTo</span><span class="hljs-params">(EdgeCost o)</span> </span>{ <span class="hljs-keyword">return</span> <span class="hljs-keyword">this</span>.cost - o.cost; } } <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">UnionFind</span> </span>{ <span class="hljs-keyword">int</span>[] parent; <span class="hljs-keyword">int</span>[] rank; <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-title">UnionFind</span><span class="hljs-params">(<span class="hljs-keyword">int</span> n)</span> </span>{ parent = <span class="hljs-keyword">new</span> <span class="hljs-keyword">int</span>[n + <span class="hljs-params">1</span>]; <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-params">0</span>; i <= n; i++) { parent[i] = i; } rank = <span class="hljs-keyword">new</span> <span class="hljs-keyword">int</span>[n + <span class="hljs-params">1</span>]; } <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-keyword">int</span> <span class="hljs-title">find</span><span class="hljs-params">(<span class="hljs-keyword">int</span> x)</span> </span>{ <span class="hljs-keyword">return</span> x == parent[x] ? x : find(parent[x]); } <span class="hljs-function"><span class="hljs-keyword">public</span> <span class="hljs-keyword">void</span> <span class="hljs-title">union</span><span class="hljs-params">(<span class="hljs-keyword">int</span> x, <span class="hljs-keyword">int</span> y)</span> </span>{ <span class="hljs-keyword">int</span> px = find(x); <span class="hljs-keyword">int</span> py = find(y); <span class="hljs-keyword">if</span> (px == py) <span class="hljs-keyword">return</span>; <span class="hljs-keyword">if</span> (rank[px] >= rank[py]) { parent[py] = px; rank[px] += rank[py]; } <span class="hljs-keyword">else</span> { parent[px] = py; rank[py] += rank[px]; } } } } ``` ``` *Pythong3 code* ``` <pre class="calibre18">``` <span class="hljs-class"><span class="hljs-keyword">class</span> <span class="hljs-title">Solution</span>:</span> <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">minCostToSupplyWater</span><span class="hljs-params">(self, n: int, wells: List[int], pipes: List[List[int]])</span> -> int:</span> union_find = {i: i <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> range(n + <span class="hljs-params">1</span>)} <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">find</span><span class="hljs-params">(x)</span>:</span> <span class="hljs-keyword">return</span> x <span class="hljs-keyword">if</span> x == union_find[x] <span class="hljs-keyword">else</span> find(union_find[x]) <span class="hljs-function"><span class="hljs-keyword">def</span> <span class="hljs-title">union</span><span class="hljs-params">(x, y)</span>:</span> px = find(x) py = find(y) union_find[px] = py graph_wells = [[cost, <span class="hljs-params">0</span>, i] <span class="hljs-keyword">for</span> i, cost <span class="hljs-keyword">in</span> enumerate(wells, <span class="hljs-params">1</span>)] graph_pipes = [[cost, i, j] <span class="hljs-keyword">for</span> i, j, cost <span class="hljs-keyword">in</span> pipes] min_costs = <span class="hljs-params">0</span> <span class="hljs-keyword">for</span> cost, x, y <span class="hljs-keyword">in</span> sorted(graph_wells + graph_pipes): <span class="hljs-keyword">if</span> find(x) == find(y): <span class="hljs-keyword">continue</span> union(x, y) min_costs += cost n -= <span class="hljs-params">1</span> <span class="hljs-keyword">if</span> n == <span class="hljs-params">0</span>: <span class="hljs-keyword">return</span> min_costs ``` ``` ## 延伸閱讀 1. [最短路徑問題](https://www.wikiwand.com/zh-hans/%E6%9C%80%E7%9F%AD%E8%B7%AF%E9%97%AE%E9%A2%98) 2. [Dijkstra算法](https://www.wikiwand.com/zh-hans/%E6%88%B4%E5%85%8B%E6%96%AF%E7%89%B9%E6%8B%89%E7%AE%97%E6%B3%95) 3. [Floyd-Warshall算法](https://www.wikiwand.com/zh-hans/Floyd-Warshall%E7%AE%97%E6%B3%95) 4. [Bellman-Ford算法](https://www.wikiwand.com/zh-hans/%E8%B4%9D%E5%B0%94%E6%9B%BC-%E7%A6%8F%E7%89%B9%E7%AE%97%E6%B3%95) 5. [Kruskal算法](https://www.wikiwand.com/zh-hans/%E5%85%8B%E9%B2%81%E6%96%AF%E5%85%8B%E5%B0%94%E6%BC%94%E7%AE%97%E6%B3%95) 6. [Prim's 算法](https://www.wikiwand.com/zh-hans/%E6%99%AE%E6%9E%97%E5%A7%86%E7%AE%97%E6%B3%95) 7. [最小生成樹](https://www.wikiwand.com/zh/%E6%9C%80%E5%B0%8F%E7%94%9F%E6%88%90%E6%A0%91)
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