Dijkstra 算法说明与实现

作者：Grey

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博客园：Dijkstra 算法说明与实现

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问题描述

问题：给定出发点，出发点到所有点的距离之和最小是多少？

注：Dijkstra 算法必须指定一个源点,每个边的权值均为非负数,求这个点到其他所有点的最短距离，到不了则为正无穷, 不能有累加和为负数的环。

题目链接见：LeetCode 743. Network Delay Time

主要思路

1. 生成一个源点到各个点的最小距离表，一开始只有一条记录，即原点到自己的最小距离为0， 源点到其他所有点的最小距离都为正无穷大

2. 从距离表中拿出没拿过记录里的最小记录，通过这个点发出的边，更新源 点到各个点的最小距离表，不断重复这一步

3. 源点到所有的点记录如果都被拿过一遍，过程停止，最小距离表得到了。

关键优化：加强堆结构说明

完整代码见：

class Solution {public static int networkDelayTime(int[][] times, int N, int K) {Graph graph = generate(times);Node from = null;for (Node n : graph.nodes.values()) {if (n.value == K) {from = n;}}HashMap map = dijkstra2(from, N);int sum = -1;for (Map.Entry entry : map.entrySet()) {if (entry.getValue() == 0) {N--;continue;}N--;if (entry.getValue() == Integer.MAX_VALUE) {return -1;} else {sum = Math.max(entry.getValue(), sum);}}// 防止出现环的形状//   int[][] times = new int[][]{{1, 2, 1}, {2, 3, 2}, {1, 3, 1}};//        int N = 3;//        int K = 2;if (N != 0) {return -1;}return sum;}public static Graph generate(int[][] times) {Graph graph = new Graph();for (int[] time : times) {int from = time[0];int to = time[1];int weight = time[2];if (!graph.nodes.containsKey(from)) {graph.nodes.put(from, new Node(from));}if (!graph.nodes.containsKey(to)) {graph.nodes.put(to, new Node(to));}Node fromNode = graph.nodes.get(from);Node toNode = graph.nodes.get(to);Edge fromToEdge = new Edge(weight, fromNode, toNode);//Edge toFromEdge = new Edge(weight, toNode, fromNode);fromNode.nexts.add(toNode);fromNode.out++;//fromNode.in++;//toNode.out++;toNode.in++;fromNode.edges.add(fromToEdge);//toNode.edges.add(toFromEdge);graph.edges.add(fromToEdge);//graph.edges.add(toFromEdge);}return graph;}public static class Graph {public HashMap nodes;public HashSet edges;public Graph() {nodes = new HashMap<>();edges = new HashSet<>();}}public static class Edge {public int weight;public Node from;public Node to;public Edge(int weight, Node from, Node to) {this.weight = weight;this.from = from;this.to = to;}}public static class Node {public int value;public int in;public int out;public ArrayList nexts;public ArrayList edges;public Node(int value) {this.value = value;in = 0;out = 0;nexts = new ArrayList<>();edges = new ArrayList<>();}}public static Node getMinNode(HashMap distanceMap, HashSet selectedNodes) {int minDistance = Integer.MAX_VALUE;Node minNode = null;for (Map.Entry entry : distanceMap.entrySet()) {Node n = entry.getKey();int distance = entry.getValue();if (!selectedNodes.contains(n) && distance < minDistance) {minDistance = distance;minNode = n;}}return minNode;}public static class NodeRecord {public Node node;public int distance;public NodeRecord(Node node, int distance) {this.node = node;this.distance = distance;}}public static class NodeHeap {private Node[] nodes; // 实际的堆结构// key 某一个node， value 上面堆中的位置private HashMap heapIndexMap;// key 某一个节点， value 从源节点出发到该节点的目前最小距离private HashMap distanceMap;private int size; // 堆上有多少个点public NodeHeap(int size) {nodes = new Node[size];heapIndexMap = new HashMap<>();distanceMap = new HashMap<>();size = 0;}public boolean isEmpty() {return size == 0;}// 有一个点叫node，现在发现了一个从源节点出发到达node的距离为distance// 判断要不要更新，如果需要的话，就更新public void addOrUpdateOrIgnore(Node node, int distance) {if (inHeap(node)) {distanceMap.put(node, Math.min(distanceMap.get(node), distance));insertHeapify(node, heapIndexMap.get(node));}if (!isEntered(node)) {nodes[size] = node;heapIndexMap.put(node, size);distanceMap.put(node, distance);insertHeapify(node, size++);}}public NodeRecord pop() {NodeRecord nodeRecord = new NodeRecord(nodes[0], distanceMap.get(nodes[0]));swap(0, size - 1);heapIndexMap.put(nodes[size - 1], -1);distanceMap.remove(nodes[size - 1]);// free C++同学还要把原本堆顶节点析构，对java同学不必nodes[size - 1] = null;heapify(0, --size);return nodeRecord;}private void insertHeapify(Node node, int index) {while (distanceMap.get(nodes[index]) < distanceMap.get(nodes[(index - 1) / 2])) {swap(index, (index - 1) / 2);index = (index - 1) / 2;}}private void heapify(int index, int size) {int left = index * 2 + 1;while (left < size) {int smallest = left + 1 < size && distanceMap.get(nodes[left + 1]) < distanceMap.get(nodes[left]) ? left + 1 : left;smallest = distanceMap.get(nodes[smallest]) < distanceMap.get(nodes[index]) ? smallest : index;if (smallest == index) {break;}swap(smallest, index);index = smallest;left = index * 2 + 1;}}private boolean isEntered(Node node) {return heapIndexMap.containsKey(node);}private boolean inHeap(Node node) {return isEntered(node) && heapIndexMap.get(node) != -1;}private void swap(int index1, int index2) {heapIndexMap.put(nodes[index1], index2);heapIndexMap.put(nodes[index2], index1);Node tmp = nodes[index1];nodes[index1] = nodes[index2];nodes[index2] = tmp;}}// 改进后的dijkstra算法// 从head出发，所有head能到达的节点，生成到达每个节点的最小路径记录并返回public static HashMap dijkstra2(Node head, int size) {NodeHeap nodeHeap = new NodeHeap(size);nodeHeap.addOrUpdateOrIgnore(head, 0);HashMap result = new HashMap<>();while (!nodeHeap.isEmpty()) {NodeRecord record = nodeHeap.pop();Node cur = record.node;int distance = record.distance;for (Edge edge : cur.edges) {nodeHeap.addOrUpdateOrIgnore(edge.to, edge.weight + distance);}result.put(cur, distance);}return result;}
}

代码说明：本题未采用题目给的二维数组的图结构，而是把二维数组转换成自己熟悉的图结构，再进行dijkstra算法。

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