# Solution For Optimize Water Distribution In A Village

Problem description:

Given a village with N houses, each house is connected to one or more other houses by pipes. Each pipe has a capacity (in liters per hour) denoted as C[i][j] where i and j are the source and destination of the pipe. The population of the village is P[i], which represents the number of people living in house i. The village has a central water treatment plant with unlimited supply, but the pipes have a limited capacity. The task is to design a water distribution system that will maximize the amount of water supplied to each house such that every house receives water at a minimum rate of one liter per hour.

Solution:

This problem is a typical minimum spanning tree (MST) problem, where the edges represent the pipes and the weights represent the capacities. The MST algorithm can be used to find the minimum capacity required to connect all the houses in the village. Then, the flow of water can be optimized using the Ford-Fulkerson algorithm.

Step 1: Build the graph

Build a graph with N nodes, where each node represents a house and each edge represents a pipe. The initial graph will have a weight matrix C[N][N] with the capacity of each edge.

Step 2: Find the minimum capacity spanning tree

Use Kruskal’s algorithm to calculate the minimum capacity spanning tree. During this process, calculate the maximum capacity of the pipes that connect the houses.

Step 3: Implement Ford-Fulkerson algorithm

Use the Ford-Fulkerson algorithm to optimize the flow of water from the central water treatment plant to each house. The idea behind Ford-Fulkerson algorithm is to increase the flow from the starting node to the ending node by finding the bottle neck capacity between them, then increase the flow of the bottleneck capacity and update the residual capacity of network, and iterate until no more augmenting paths are found. In this problem, the starting node is the central water treatment plant and the ending node is every house in the village. The residual capacity of the network is calculated as the difference between the total capacity and the current flow.

Step 4: Calculate the maximum flow

Calculate the maximum flow of water that can be distributed to each house while maintaining a minimum flow rate of one liter per hour.

Time complexity of the solution:

The time complexity of Kruskal’s algorithm is O(ElogE), where E is the number of edges in the graph. The time complexity of Ford-Fulkerson algorithm is O(Ef), where f is the maximum flow. Therefore, the time complexity of the solution is O(ElogE + Ef).

Code:

The code for the solution of this problem can be found on the LeetCode website.

## Step by Step Implementation For Optimize Water Distribution In A Village

There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes. For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are bidirectional. Find the minimum total cost to supply water to all houses. class Solution { public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) { // create a graph with n+1 nodes (0 node for the water source) int[][] graph = new int[n+1][n+1]; for (int[] pipe : pipes) { // connect the two houses together with the cost graph[pipe[0]][pipe[1]] = pipe[2]; graph[pipe[1]][pipe[0]] = pipe[2]; } // create an array to store the minimum cost to connect to the water source int[] minCost = new int[n+1]; // initialize the minCost array with the cost of each well for (int i = 0; i < n; i++) { minCost[i+1] = wells[i]; } // use prim's algorithm to find the minimum cost spanning tree boolean[] visited = new boolean[n+1]; PriorityQueuepq = new PriorityQueue<>(n, (a,b) -> a[2]-b[2]); // start from node 1 (the first house) pq.add(new int[]{1,1,0}); while (!pq.isEmpty()) { int[] curr = pq.poll(); int currNode = curr[1]; if (visited[currNode]) continue; visited[currNode] = true; // update the minCost of the current node minCost[currNode] = curr[2]; // add all the neighbors of the current node to the priority queue for (int i = 1; i <= n; i++) { if (graph[currNode][i] > 0 && !visited[i]) { pq.add(new int[]{currNode, i, graph[currNode][i]}); } } } // sum up the minCost array to get the total cost int totalCost = 0; for (int cost : minCost) { totalCost += cost; } return totalCost; } }

There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes. For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are always bidirectional. Find the minimum total cost to supply water to all houses. Example 1: Input: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]] Output: 3 Explanation: 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. Constraints: 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]

There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes. For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are bidirectional. Find the minimum total cost to supply water to all houses. var minCost = function(n, wells, pipes) { // your code goes here };

There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes. For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are always bidirectional. Find the minimum total cost to supply water to all houses. Example 1: Input: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]] Output: 3 Explanation: 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. Constraints: 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]

There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes. For each house i, we can either build a well inside it directly with cost wells[i], or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes, where each pipes[i] = [house1, house2, cost] represents the cost to connect house1 and house2 together using a pipe. Connections are bidirectional. Find the minimum total cost to supply water to all houses. Solution: We can use a greedy algorithm to solve this problem. First, we sort the houses by their distance from the well. Then, we connect each house to the nearest well, and update the cost accordingly. Finally, we return the total cost. public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) { // sort houses by distance from the well int[] dist = new int[n]; for (int i = 0; i < n; i++) { dist[i] = wells[i]; } Arrays.sort(dist); // connect each house to the nearest well int cost = 0; for (int i = 0; i < n; i++) { cost += dist[i]; // update the cost of the pipes for (int j = 0; j < pipes.length; j++) { if (pipes[j][0] == dist[i]) { pipes[j][2] += cost; } } } return cost; }