概述
- 一般深度优先搜索问题 => 一般图 + 矩阵(二维数组)
- DFS + 涂色标记(避免节点重复访问)
- 搜索节点如何移动
- 同时能用 DFS 或 BFS 求解的图或矩阵问题，优先考虑 BFS
图论
- 图论（Graph theory），是组合数学分支，和其他数学分支如群论、矩阵论、拓扑学有着密切关系。
- 图是图论的主要研究对象。图是由若干给定的顶点及连接两顶点的边所构成的图形，这种图形通常用来描述某些事物之间的某种特定关系。顶点用于代表事物，连接两顶点的边则用于表示两个事物间具有这种关系。
- 图G(V, E) => 由一系列顶点(Vertices)和边(Edges)连接而成
分类
- 有向图 => 有向边组成的图
- 有向边e => e = (u, v) => u 为边的起始顶点，v 为边的终止顶点
- 有向无环图(DAG) => 无法从某个顶点出发经过若干条边回到该点的有向图 => 只存在于有向图
- 无向图
- 带权图 => 边有权值
图的表示
- 邻接矩阵 => Adjacency Matrix => 容易变为稀疏矩阵，开辟空间巨大，浪费内存
- 无向图节点 i 的度 => 第 i 行或第 i 列非 0 个数
- 有向图节点 i 的出度 => 第 i 行(行表示起点)非 0 个数
- 有向图节点 i 的入度 => 第 i 列(列表示终点)非 0 个数
- 邻接表 => Adjacency List => 描述节点与相邻边的映射，围绕数据结构去建模
- 链表
- Map<Node, List<Node>>
- 出度方便计算，入度需要遍历，或者创建一个逆邻接表，用来表示入度的节点
图搜索
DFS
- 时间复杂度
- 邻接表 => O(|V| + |E|) => 和点边都有关系 => 描述点与相邻边的映射
- 邻接矩阵 => O(|V|^2) => 和点有关，和边无关
- 空间复杂度 => O(|V|)
BFS
- 时间复杂度
- 邻接表 => O(|V| + |E|) => 和点边都有关系 => 描述点与相邻边的映射
- 邻接矩阵 => O(|V|^2) => 和点有关，和边无关
- 空间复杂度 => O(|V|)
一般 DFS 问题
图中 DFS
public class UndirectedGraph {
// 邻接表 => key: 当前节点 value: 邻接节点集合
private final Map<Node, List<Node>> adjacencyList;
// 是否被访问涂色过
private final boolean[] marked;
// 统计连通分量
private int connectComponentCount;
public UndirectedGraph(int vertexCount) {
this.adjacencyList = new HashMap<>();
this.marked = new boolean[vertexCount];
this.connectComponentCount = 0;
}
// Method
public void addVertex(Node v) {
if (!adjacencyList.containsKey(v)) {
adjacencyList.put(v, new ArrayList<>());
}
}
public void addEdge(Node u, Node v) {
if (!adjacencyList.containsKey(u)) {
addVertex(u);
}
if (!adjacencyList.containsKey(v)) {
addVertex(v);
}
adjacencyList.get(u).add(v);
adjacencyList.get(v).add(u);
}
public void printGraph() {
for (Node key : adjacencyList.keySet()) {
System.out.println(key.getNo() + ": " + adjacencyList.get(key).stream().map(Node::getNo).map(String::valueOf).collect(Collectors.joining(",", "[", "]")));
}
}
// DFS 模板 => 遍历全图
public void dfsInGraph() {
Arrays.fill(marked, false);
for (Node startNode : adjacencyList.keySet()) {
if (!marked[startNode.getNo()]) {
helper(startNode);
connectComponentCount++;
}
}
}
// DFS 模板 => 以 start 节点为起点在某一连通分量上 DFS
public void dfsInComponent(Node start) {
Arrays.fill(marked, false);
helper(start);
}
// DFS helper 函数
public void helper(Node start) {
marked[start.getNo()] = true;
System.out.println("DFS current node: " + start.getNo() + " value: " + start.getValue());
for (Node adjNode : adjacencyList.get(start)) {
if (!marked[adjNode.getNo()]) {
// 没有被访问过
helper(adjNode);
}
}
}
public static void main(String[] args) {
// 0 - 1 - 4
// | | /
// 2 - 3
UndirectedGraph undirectedGraph = new UndirectedGraph(7);
Node zero = new Node(0, 1);
Node one = new Node(1, 2);
Node two = new Node(2, 3);
Node three = new Node(3, 4);
Node four = new Node(4, 5);
Node five = new Node(5, 6);
Node six = new Node(6, 7);
undirectedGraph.addEdge(zero, one);
undirectedGraph.addEdge(zero, two);
undirectedGraph.addEdge(one, four);
undirectedGraph.addEdge(one, three);
undirectedGraph.addEdge(three, two);
undirectedGraph.addEdge(three, four);
undirectedGraph.addEdge(five, six);
undirectedGraph.printGraph();
System.out.println("Starting from node 0");
undirectedGraph.dfsInComponent(zero);
System.out.println("Starting from node 1");
undirectedGraph.dfsInComponent(one);
System.out.println("Traversal graph");
undirectedGraph.dfsInGraph();
System.out.println(undirectedGraph.connectComponentCount);
}
}
class Node {
int no;
int value;
public Node(int no, int value) {
this.no = no;
this.value = value;
}
public int getNo() {
return no;
}
public void setNo(int no) {
this.no = no;
}
public int getValue() {
return value;
}
public void setValue(int value) {
this.value = value;
}
}
模板
import java.util.HashMap;
class DFSInGraph {
public void dfsInGraph(int nodeNum, int[][] edges, int[][] adjacencyMatrix) {
// 0. 邻接表 => Map<Node, List<Node>> => key: current node value: adjacency nodes
// 邻接矩阵 => int[][] => 索引值代表节点(事物)，(x, y) 的值表示是否相邻
// 1. check input
if (adjacencyMatrix == null || adjacencyMatrix.length == 0 || adjacencyMatrix[0] == null || adjacencyMatrix[0].length == 0) {
return;
}
// 2. 边的数组 => 构建邻接表 => 非必须
Map<Integer, List<Integer>> adjacencyList = new HashMap<>();
for (int i = 0; i < nodeNum; i++) {
adjacencyList.put(i, new ArrayList<>());
}
for (int i = 0; i < edges.length; i++) {
int u = edges[i][0];
int v = edges[i][1];
adjacencyList.get(u).add(v);
adjacencyList.get(v).add(u);
}
// 3. marked
boolean[] visited = new boolean[adjacencyMatrix.length];
int connectComponentCount = 0;
// 4. traversal adjacency matrix
for (int i = 0; i < adjacencyMatrix.length; i++) {
if (!visited[i]) {
helper(adjacencyMatrix, adjacencyList, visited, i);
connectComponentCount++;
}
}
// 4. traversal adjacency list
for (int node : adjacencyList.keySet()) {
if (visited[node]) {
helper(adjacencyMatrix, adjacencyList, visited, node);
connectComponentCount++;
}
}
}
private void helper(int[][] adjacencyMatrix, Map<Integer, List<Integer>> adjacencyList, boolean[] visited, int start) {
// marked
visited[start] = true;
// traversal adjacency node in adjacencyMatrix
for (int i = 0; i < adjacencyMatrix[start].length; i++) {
if (specialCondition && !visited[i]) {
helper(adjacencyMatrix, visited, i);
}
}
// traversal adjacency node in adjacencyList
for (int item : adjacencyList.get(start)) {
if (specialCondition && !visited[item]) {
helper(adjacencyMatrix, adjacencyList, visited, item);
}
}
}
}
二维问题
- 矩阵/网格
分类
- 迷宫问题 => 二维数组上的回溯法
- 岛屿问题 => 连通分量
特点
- 二维数组/矩阵问题有区域限制和移动方向限制
- 能移动的方向集合即为邻接节点集合
- 题目有特定条件限制
重点
- DFS + 涂色标记(标记已访问过的位置)
- 剪枝
二维空间移动
坐标 (x, y) => 以左下角为原点 => 注意移动边界限制
- 向上移动 => (x, y + 1)
- 向下移动 => (x, y - 1)
- 向左移动 => (x - 1, y)
- 向右移动 => (x + 1, y)
- 向右上移动 => (x + 1, y + 1)
- 向左上移动 => (x - 1, y + 1)
- 向左下移动 => (x - 1, y - 1)
- 向右下移动 => (x + 1, y - 1)
// version1
final int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
for (int[] offset : directions) {
Point next = new Point(cur.x + offset[0], current.y + offset[1]);
}
// version2
int[] dx = {1, 0, -1, 0};
int[] dy = {0, 1, 0, -1};
for (int i = 0; i < 4; i++) {
Point next = new Point(cur.x + dx[i], cur.y + dy[i]);
}
模板
考虑问题
- specialCondition 是什么？=> 开始能做 DFS 是否有特殊条件?
- pruning
- how to move
- is need backtracking
class DFSInMatrix {
public void dfsInMatrix(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) {
return;
}
// marked
int m = matrix.length;
int n = matrix[0].length;
boolean[][] visited = new boolean[m][n];
int connectedComponentCount = 0;
// traversal
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (specialCondition && !visited[i][j]) {
dfs(matrix, visited, i, j);
connectedComponentCount++;
}
}
}
}
private void dfs(int[][] matrix, boolean[][] visited, int x, int y) {
// pruning => 此处的 pruning 也可以放置在 move 中
if (condition) {
return;
}
// marked
visited[x][y] = true;
// move
int[] dx = {1, 0, -1, 0};
int[] dy = {0, 1, 0, -1};
for (int i = 0; i < 4; i++) {
int newX = x + dx[i];
int newY = y + dy[i];
// 此处可以进行 pruning => if (checkRange(matrix, newX, newY) && !visited[newX][newY] && specialCondition)
if (checkRange(matrix, newX, newY) && !visited[newX][newY]) {
dfs(matrix, visited, newX, newY);
}
}
}
private boolean checkRange(int[][] matrix, int x, int y) {
return x >= 0 && x < matrix.length && y >= 0 && y < matrix[0].length;
}
}
知识点
- 度 => 图中的节点有几个边就有几个度
- 连通分量 => 一个图有几个互不相干的部分
- 在图中的邻接矩阵中索引表示事物，在二维矩阵中矩阵中的点表示事物
网友评论