16：DFS和BFS

搜索算法==》在图中找出从一个顶点出发，到另一个顶点的路径

public class Graph {  //无向图
    private int v;  //顶点的个数
    private LinkedList<Integer> adj[];  //邻接表
    
    public Graph(int v) {
        this.v = v;
        adj = new LinkedList[v];
        for (int i = 0; i < v; ++i) {
            adj[i] = new LinkedList<>();
        }
    }

    public void addEdge(int s, int t) {  //无向图一条边存两次
        adj[s].add(t);
        adj[t].add(s);
    }
}

==>BFS

s 表示起始顶点，t 表示终止顶点。我们搜索一条从 s 到 t 的路径。实际上，这样求得的路径就是从 s 到 t 的最短路径。

public void bfs(int s, int t) {
    if (s == t) return;
    boolean[] visited = new boolean[v];
    visited[s] = true;
    Queue<Integer> queue = new LinkedList<>();
    queue.add(s);
    int[] prev = new int[v];
    for (int i = 0; i < v; ++i) {
        prev[i] = -1;
    }
    while (queue.size() != 0) {
        int w = queue.poll();
        for (int i = 0; i < adj[w].size(); ++i) {
            int q = adj[w].get(i);
            if (!visited[q]) {
                prev[q] = w;
                if (q == t) {
                    print(prev, s, t);
                    return;
                }
                visited[q] = true;
                queue.add(q);
            }
        }
    }
}

private void print(int[] prev, int s, int t) {  //递归打印s->t的路径
    if (prev[t] != -1 && t != s) {
        print(prev, s, prev[t]);
    }
    System.out.print(t + " ");
}

==》DFS

public class Dfs {
    boolean found = false;  //全局变量或者类成员变量
    
    public void dfs(int s, int t) {
        found = false;
        boolean[] visited = new boolean[v];
        int[] prev = new int[v];
        for (int i = 0; i < v; ++i) {
            prev[i] = -1;
        }
        recurDfs(s, t, visited, prev);
        print(prev, s, t);
    }

    private void recurDfs(int w, int t, boolean[] visited, int[] prev) {
        if (found == true) return;
        visited[w] = true;
        if (w == t) {
            found = true;
            return;
        }
        for (int i = 0; i < adj[w].size(); ++i) {
            int q = adj[w].get(i);
            if (!visited[q]) {
                prev[q] = w;
                recurDfs(q, t, visited, prev);
            }
        }
    }
}