给我用Java 写出一个贪心算法

时间: 2023-05-13 13:07:44 浏览: 44
我可以给你写一个贪心算法的Java程序: public class GreedyAlgorithm { public static void main(String[] args) { int[] arr = {7, 10, 12, 15, 20}; int sum = 0; int i = 0; while (i < arr.length) { if (arr[i] + sum <= 20) { sum += arr[i]; } i++; } System.out.println("最大和为:" + sum); } }
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用java写一个贪心算法

好的,以下是一个使用 Java 编写的贪心算法示例: ``` import java.util.Arrays; public class GreedyAlgorithm { public static void main(String[] args) { int[] coins = {1, 5, 10, 25}; // 硬币面值 int amount = 63; // 需要凑出的金额 int[] usedCoins = new int[coins.length]; // 记录每种硬币使用的数量 // 从大到小遍历硬币面值 for (int i = coins.length - 1; i >= 0; i--) { int coin = coins[i]; while (amount >= coin) { // 如果金额大于等于当前硬币面值 amount -= coin; // 减去当前硬币面值 usedCoins[i]++; // 记录当前硬币使用数量 } } // 输出结果 System.out.println("需要的硬币数量:" + Arrays.stream(usedCoins).sum()); for (int i = 0; i < coins.length; i++) { System.out.println(coins[i] + " 元硬币使用了 " + usedCoins[i] + " 个"); } } } ``` 这个程序实现了一个简单的贪心算法,用最少的硬币凑出指定的金额。

用java写一个最短路径算法

### 回答1: 您可以使用 Dijkstra 算法来解决最短路径问题。该算法基于贪心策略,通过不断地选择未确定的最短路径节点来实现。在 Java 中,您可以使用优先队列来维护未确定的节点,从而确定最短路径。 以下是 Java 代码的简单实现: ``` import java.util.*; class Dijkstra { private int dist[]; private Set<Integer> settled; private PriorityQueue<Node> pq; private int V; // Number of vertices List<List<Node>> adj; public Dijkstra(int V) { this.V = V; dist = new int[V]; settled = new HashSet<Integer>(); pq = new PriorityQueue<Node>(V, new Node()); } // Function for Dijkstra's Algorithm public void dijkstra(List<List<Node>> adj, int src) { this.adj = adj; for (int i = 0; i < V; i++) dist[i] = Integer.MAX_VALUE; // Add source node to the priority queue pq.add(new Node(src, 0)); // Distance to the source is 0 dist[src] = 0; while (settled.size() != V) { // remove the minimum distance node // from the priority queue int u = pq.remove().node; // adding the node whose distance is // finalized settled.add(u); e_Neighbours(u); } } // Function to process all the neighbours // of the passed node private void e_Neighbours(int u) { int edgeDistance = -1; int newDistance = -1; // All the neighbors of v for (int i = 0; i < adj.get(u).size(); i++) { Node v = adj.get(u).get(i); // If current node hasn't already been processed if (!settled.contains(v.node)) { edgeDistance = v.cost; newDistance = dist[u] + edgeDistance; // If new distance is cheaper in cost if (newDistance < dist[v.node]) dist[v.node] = newDistance; // Add the current node to the queue pq.add(new Node(v.node, dist[v.node])); } } } // Driver code public static void main(String arg[]) { int V = 5; int source = ### 回答2: 最短路径算法是一种计算图中两个节点之间最短路径的算法。常见的最短路径算法有Dijkstra算法和Floyd-Warshall算法。 在Java中,我们可以使用Dijkstra算法来实现最短路径的计算。首先,我们需要定义一个图的数据结构,用来存储节点以及节点之间的边。可以使用邻接矩阵或邻接表来表示图。接下来,我们可以使用Java的优先队列来实现Dijkstra算法。 算法步骤如下: 1. 创建一个优先队列和一个距离数组,用于存储节点到起始节点的最短距离。将起始节点的距离设置为0,其他节点的距离设置为无限大,并将起始节点加入优先队列。 2. 循环直到优先队列为空: a. 弹出队列中距离最小的节点; b. 遍历该节点的邻接节点,如果通过当前节点到达邻接节点的距离小于目前存储的距离,则更新距离数组,并将邻接节点加入优先队列。 3. 最终,距离数组中存储了每个节点到起始节点的最短路径距离。 以下是一个简单的使用Java实现Dijkstra算法的示例代码: ```java import java.util.*; public class Dijkstra { public static void main(String[] args) { int[][] graph = { { 0, 4, 0, 0, 0, 0, 0, 8, 0 }, { 4, 0, 8, 0, 0, 0, 0, 11, 0 }, { 0, 8, 0, 7, 0, 4, 0, 0, 2 }, { 0, 0, 7, 0, 9, 14, 0, 0, 0 }, { 0, 0, 0, 9, 0, 10, 0, 0, 0 }, { 0, 0, 4, 14, 10, 0, 2, 0, 0 }, { 0, 0, 0, 0, 0, 2, 0, 1, 6 }, { 8, 11, 0, 0, 0, 0, 1, 0, 7 }, { 0, 0, 2, 0, 0, 0, 6, 7, 0 } }; int[] shortestDistances = dijkstra(graph, 0); System.out.println("起始节点到其他节点的最短路径距离:"); for (int i = 0; i < shortestDistances.length; i++) { System.out.println("节点 " + i + " 的最短路径距离为:" + shortestDistances[i]); } } public static int[] dijkstra(int[][] graph, int start) { int n = graph.length; int[] distances = new int[n]; Arrays.fill(distances, Integer.MAX_VALUE); distances[start] = 0; PriorityQueue<Node> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a.distance)); pq.add(new Node(start, 0)); while (!pq.isEmpty()) { Node current = pq.poll(); int currentNode = current.node; int currentDistance = current.distance; if (currentDistance > distances[currentNode]) { continue; } for (int i = 0; i < n; i++) { if (graph[currentNode][i] != 0) { int newDistance = currentDistance + graph[currentNode][i]; if (newDistance < distances[i]) { distances[i] = newDistance; pq.add(new Node(i, newDistance)); } } } } return distances; } static class Node { int node; int distance; public Node(int node, int distance) { this.node = node; this.distance = distance; } } } ``` 该示例代码使用邻接矩阵表示图,并计算了起始节点到其他节点的最短路径距离。 ### 回答3: 最短路径算法是一种用于在图中找到两个节点之间最短路径的算法。其中最著名的算法之一是Dijkstra算法。下面是用Java语言编写Dijkstra算法的一个简单实现: ```java import java.util.*; public class ShortestPathAlgorithm { public void dijkstra(int[][] graph, int startVertex) { int numVertices = graph[0].length; int[] shortestDistances = new int[numVertices]; boolean[] visited = new boolean[numVertices]; for (int i = 0; i < numVertices; i++) { shortestDistances[i] = Integer.MAX_VALUE; visited[i] = false; } shortestDistances[startVertex] = 0; for (int i = 0; i < numVertices - 1; i++) { int minDistanceVertex = findMinDistanceVertex(shortestDistances, visited); visited[minDistanceVertex] = true; for (int j = 0; j < numVertices; j++) { if (!visited[j] && graph[minDistanceVertex][j] != 0 && shortestDistances[minDistanceVertex] != Integer.MAX_VALUE && shortestDistances[minDistanceVertex] + graph[minDistanceVertex][j] < shortestDistances[j]) { shortestDistances[j] = shortestDistances[minDistanceVertex] + graph[minDistanceVertex][j]; } } } // 输出最短路径 for (int i = 0; i < numVertices; i++) { System.out.println("从起始节点到节点 " + i + " 的最短路径距离为: " + shortestDistances[i]); } } private int findMinDistanceVertex(int[] shortestDistances, boolean[] visited) { int minDistance = Integer.MAX_VALUE; int minDistanceVertex = -1; for (int i = 0; i < shortestDistances.length; i++) { if (!visited[i] && shortestDistances[i] < minDistance) { minDistance = shortestDistances[i]; minDistanceVertex = i; } } return minDistanceVertex; } public static void main(String[] args) { int[][] graph = { {0, 4, 0, 0, 0, 0, 0, 8, 0}, {4, 0, 8, 0, 0, 0, 0, 11, 0}, {0, 8, 0, 7, 0, 4, 0, 0, 2}, {0, 0, 7, 0, 9, 14, 0, 0, 0}, {0, 0, 0, 9, 0, 10, 0, 0, 0}, {0, 0, 4, 14, 10, 0, 2, 0, 0}, {0, 0, 0, 0, 0, 2, 0, 1, 6}, {8, 11, 0, 0, 0, 0, 1, 0, 7}, {0, 0, 2, 0, 0, 0, 6, 7, 0} }; ShortestPathAlgorithm shortestPathAlgorithm = new ShortestPathAlgorithm(); shortestPathAlgorithm.dijkstra(graph, 0); } } ``` 在这个实现中,我们首先创建一个数组`shortestDistances`,其中存储着从起始节点到各个节点的最短路径距离。我们还创建一个布尔类型的数组`visited`,用于标记已访问过的节点。 然后,我们将起始节点的最短路径距离设置为0,并遍历图中的所有节点。对于每个节点,我们选择一个最短路径距离最小的节点,并将其标记为已访问。之后,我们检查当前节点与其相邻节点之间的路径是否产生了更短的路径,如果是的话,我们更新这些节点的最短路径距离。 最后,我们打印出从起始节点到每个节点的最短路径距离。 上述代码中给出了一个示例图,你可以修改此图或定义自己的图来测试该算法的功能。

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1444. Elephpotamus Time limit: 0.5 second Memory limit: 64 MB Harry Potter is taking an examination in Care for Magical Creatures. His task is to feed a dwarf elephpotamus. Harry remembers that elephpotamuses are very straightforward and imperturbable. In fact, they are so straightforward that always move along a straight line and they are so imperturbable that only move when attracted by something really tasty. In addition, if an elephpotamus stumbles into a chain of its own footprints, it falls into a stupor and refuses to go anywhere. According to Hagrid, elephpotamuses usually get back home moving along their footprints. This is why they never cross them, otherwise they may get lost. When an elephpotamus sees its footprints, it tries to remember in detail all its movements since leaving home (this is also the reason why they move along straight lines only, this way it is easier to memorize). Basing on this information, the animal calculates in which direction its burrow is situated, then turns and goes straight to it. It takes some (rather large) time for an elephpotamus to perform these calculations. And what some ignoramuses recognize as a stupor is in fact a demonstration of outstanding calculating abilities of this wonderful, though a bit slow-witted creature. Elephpotamuses' favorite dainty is elephant pumpkins, and some of such pumpkins grow on the lawn where Harry is to take his exam. At the start of the exam, Hagrid will drag the elephpotamus to one of the pumpkins. Having fed the animal with a pumpkin, Harry can direct it to any of the remaining pumpkins. In order to pass the exam, Harry must lead the elephpotamus so that it eats as many pumpkins as possible before it comes across its footprints. Input The first input line contains the number of pumpkins on the lawn N (3 ≤ N ≤ 30000). The pumpkins are numbered from 1 to N, the number one being assigned to the pumpkin to which the animal is brought at the start of the trial. In the next N lines, the coordinates of the pumpkins are given in the order corresponding to their numbers. All the coordinates are integers in the range from −1000 to 1000. It is guaranteed that there are no two pumpkins at the same location and there is no straight line passing through all the pumpkins. Output In the first line write the maximal number K of pumpkins that can be fed to the elephpotamus. In the next K lines, output the order in which the animal will eat them, giving one number in a line. The first number in this sequence must always be 1.写一段Java完成此目的

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