标签:integer image final 计算 null ted 数组 iterable push
无向图
我们用邻接图来表示图
具体实现的代码
public class Graph { private final int V; private int E; private Bag<Integer>[] adj; public Graph(int V){ this.V = V; E = 0; adj = (Bag<Integer>[]) new Bag[V]; for (int i = 0; i < V; i++) { adj[i] = new Bag<Integer>(); } } public int V(){ return V; } public int E(){ return E; } public void addEdge(int v , int w){ //将v w 连接起来 adj[v].add(w); adj[w].add(v); E++; } public Iterable<Integer> adj(int v){ return adj[v]; } //计算v 的度数 public static int degree(Graph G , int v){ int degree = 0; for (Integer integer : G.adj(v)) { degree++; } return degree; } public static int maxDegree(Graph G){ int max = 0; for (int i = 0; i < G.V(); i++) { if (degree(G,i) > max)max = degree(G,i); } return max; } //计算所有顶点的平均度数 public static double avgDegree(Graph G){ return 2.0*G.E()/ G.V(); } //计算所有自环的个数 public static int numberOfSelfLoops(Graph G){ int count = 0; for (int v = 0; v < G.V(); v++) { for (Integer item : G.adj[v]) { if (item == v)count++; } } return count/2; } public String toString(){ String s = V + "vertices," + E + "edges\n"; for (int v = 0; v < V; v++) { s += v + ":"; for (Integer w : adj(v)) { s += w + " "; } s += "\n"; } return s; } }
这里注意泛型数组的创建 adj = (Bag<Integer>[]) new Bag[V];
DFS
public class DepthFirstPaths { private boolean[] marked; private int[] edgeTo; private final int s; // 起点 public DepthFirstPaths(Graph G ,int s){ this.s = s; marked = new boolean[G.V()]; edgeTo = new int[G.V()]; dfs(G,s); } private void dfs(Graph G , int v){ marked[v] = true; for (Integer w : G.adj(v)) { if (!marked[w]){ edgeTo[w] = v; dfs(G,w); } } } public boolean hasPathTo(int v){ return marked[v]; } public Iterable<Integer> pathTo(int v){ if (marked[v] == false)return null; Stack<Integer> path = new Stack<>(); for(int x = v;x !=s; x = edgeTo[x]){ path.push(x); } path.push(s); return path; }
dfs的主要思想:用递归的方法,先把一条路走到底。其实dfs运用了java系统自带的栈。
BFS
public class BreadthFirstPaths { private boolean[] marked; private int[] edgeTo; private final int s; private int[] dis; public BreadthFirstPaths(Graph G , int s){ marked = new boolean[G.V()]; edgeTo = new int[G.V()]; this.s = s; dis = new int[G.V()]; bfs(G , s); } private void bfs(Graph G , int s){ Queue<Integer> queue = new Queue<Integer>(); marked[s] = true; queue.enqueue(s); while(!queue.isEmpty()){ int v = queue.dequeue(); for (Integer w : G.adj(v)) { if (!marked[w]){ edgeTo[w] = v; marked[w] = true; dis[w] = dis[v] + 1; queue.enqueue(w); } } } } public int distance(int w){ return dis[w]; } public boolean hasPathTo(int v){ return marked[v]; } public Iterable<Integer> pathTo(int v){ if (!marked[v] )return null; Stack<Integer> path = new Stack<>(); for (int w = v;w !=s;w = edgeTo[w]){ path.push(w); } path.push(s); return path; }
bfs 是一层一层搜索,一层搜索完再进入到下一层,运用了queue先进先出的特质。
同时bfs还能用于搜索最短路径。
CC(Connected Components)
这里的连通分量和之前union find 并查集中的一样。
但是这里我们用dfs来查找有多少个cc
public class CC { private boolean[] marked; private int count; private int[] id; public CC(Graph G){ marked = new boolean[G.V()]; count = 0; int[] id = new int[G.V()]; for (int i = 0; i < G.V(); i++) { if (!marked[i]){ dfs(G,i); count++; } } } private void dfs(Graph G , int v){ marked[v] = true; id[v] = count; for (Integer w : G.adj(v)) { if (!marked[w]){ dfs(G,w); } } } }
一次dfs 可以把 与 v在一条路径上的点全部打通。
标签:integer image final 计算 null ted 数组 iterable push
原文地址:https://www.cnblogs.com/JasonJ/p/14496346.html