Dijkstra 算法说明与实现
作者：Grey
原文地址：
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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
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
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
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<Node, Integer> map = dijkstra2(from, N);
int sum = -1;
for (Map.Entry<Node, Integer> 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<Integer, Node> nodes;
public HashSet<Edge> 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<Node> nexts;
public ArrayList<Edge> edges;
public Node(int value) {
this.value = value;
in = 0;
out = 0;
nexts = new ArrayList<>();
edges = new ArrayList<>();
}
}
public static Node getMinNode(HashMap<Node, Integer> distanceMap, HashSet<Node> selectedNodes) {
int minDistance = Integer.MAX_VALUE;
Node minNode = null;
for (Map.Entry<Node, Integer> 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<Node, Integer> 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<Node, Integer> 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<Node, Integer> dijkstra2(Node head, int size) {
NodeHeap nodeHeap = new NodeHeap(size);
nodeHeap.addOrUpdateOrIgnore(head, 0);
HashMap<Node, Integer> 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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