精进DFS和BFS，按照流程来，秒杀
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DFS，BFS市面上都有很多的代码和视频，很多写的不明就里。其实一个是（基于栈的）递归，一个是队列的遍历。可以参照以下流程。顺一遍下来，就可以解决各种问题了。
例题1：DFS遍历
代码：
import java.util.Scanner;
import java.util.Stack;
public class reWriteDFS
public static int time = 0;
public static void main(String[] args)
Scanner cin = new Scanner(System.in);
int n = cin.nextInt();//n个节点
int[][] M = new int[n][n];
for (int i=0;i<n;i++)
int id = cin.nextInt();
int degree = cin.nextInt();
for (int j=0;j<degree;j++)
int neighbour = cin.nextInt();
M[id-1][neighbour-1] = 1;
Stack<Integer> S = new Stack<>();
int[] d = new int[n];
int[] f = new int[n];
int[] process = new int[n];
dfs_init(d, f, S, M, n, process);
for (int i:d)
System.out.print(i+" ");
System.out.println();
System.out.println();
for (int i:f)
System.out.print(i+" ");
public static void dfs_init(int[] d, int[] f, Stack<Integer> S, int[][] M, int n, int[] process)
S.push(0);
time++;
process[0] = 1;
d[0] = time;
dfs(0, d, f, S, M, n, process);
public static void dfs(int u, int[] d, int[] f, Stack<Integer> S, int[][] M, int n, int[] process)
int tag = hasNext(u, M, process);
if(tag ==-1)
time++;
f[u] = time;
S.pop();
if(S.size()==0)
return;
else
int v = S.peek();
dfs(v, d, f, S, M, n, process);
else
S.push(tag);
time++;
d[tag] = time;
process[tag] = 1;
dfs(tag, d, f, S, M, n, process);
public static int hasNext(int u, int[][] M, int[] process)
for (int i=0;i<process.length;i++)
if(M[u][i]==1 && process[i]==0)
return i;
return -1;
例题2：用DFS求连通块
代码：
import java.util.Stack;
public class uva572
public static int id = 0;
static class Position
int x;
int y;
Position()
public Position(int x, int y)
this.x = x;
this.y = y;
public static void main(String[] args)
String map[] = new String[5];
map[0] = "****@";
map[1] = "*@@*@";
map[2] = "*@**@";
map[3] = "@@@*@";
map[4] = "@@**@";
int[][] group = new int[5][5];
int[][] process = new int[5][5];
Stack<Position> S = new Stack<>();
for (int i=0;i<5;i++)
for (int j=0;j<5;j++)
//对每一个元素进行判断，如果是*，process[i][j]就设置为-1，如果是@就设置为1
if(map[i].charAt(j)=='*')
process[i][j] = -1;
else if(map[i].charAt(j)=='@' && process[i][j] == 0)
Position u = new Position(i,j);
S.push(u);
process[i][j] = 1;
id++;
group[i][j] = id;
dfs(u, process, group, S, map);
else if(map[i].charAt(j)=='@' && process[i][j] == 1)
continue;
System.out.println(id);
public static void dfs(Position u, int[][] process, int[][] group, Stack<Position> S, String map[])
Position v = hasNext(u, map, process);
if(v.x==-1 && v.y==-1)
S.pop();
if (S.size()==0)
return;
else
Position tag = S.peek();
dfs(tag, process, group, S, map);
else
S.push(v);
process[v.x][v.y] = 1;
group[v.x][v.y] = id;
dfs(v, process, group, S, map);
public static Position hasNext(Position u, String map[], int[][] process)
for (int i=-1;i<=1;i++)
for (int j=-1;j<=1;j++)
int x_new = u.x+i;
int y_new = u.y+j;
if(x_new<5 && x_new>=0 && y_new<5 && y_new>=0)//下标符合，不符合都不用看
if(map[x_new].charAt(y_new)=='*')
process[x_new][y_new] = -1;
else if(process[x_new][y_new] == 0)//只要找到了还没有被处理过的@，那就返回他的位置
return new Position(x_new, y_new);
return new Position(-1, -1);
例题3：BFS遍历
代码：
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Scanner;
public class rewriteBFS
public static int time = 0;
public static void main(String[] args)
Scanner cin = new Scanner(System.in);
int n = cin.nextInt();
int[] process = new int[n];
ArrayDeque<Integer> Q = new ArrayDeque<>();
int[] d = new int[n];
int[] f = new int[n];
//分别使用两种存储方式进行bfs
/* //1.连接矩阵
int[][] M = new int[n][n];
for (int i=0;i<n;i++)
int id = cin.nextInt();
int degree = cin.nextInt();
for (int j=0;j<degree;j++)
int neighbour = cin.nextInt();
M[id-1][neighbour-1] = 1;
bfs_init1(M, process, Q, d, f);*/
ArrayList<Integer> al[] = new ArrayList[n];
for (int i=0;i<n;i++)
al[i] = new ArrayList<>();
for (int i=0;i<n;i++)
int id = cin.nextInt();
int degree = cin.nextInt();
for (int j=0;j<degree;j++)
int neighbour = cin.nextInt();
al[id-1].add(neighbour-1);
bfs_init2(al, process, Q, d, f);
for (int i:d)
System.out.print(i+" ");
System.out.println();
for (int i:f)
System.out.print(i+" ");
public static void bfs_init1(int[][] M, int[] process, ArrayDeque<Integer> Q, int[] d, int[] f)
Q.addLast(0);
process[0] = 1;
bfs1(0, M, process, Q, d, f);
public static void bfs1(int u, int[][] M, int[] process, ArrayDeque<Integer> Q, int[] d, int[] f)
for (int i=0;i<process.length;i++)
if(M[u][i]==1 && process[i]==0)
Q.addLast(i);
process[i] = 1;
time++;
d[i] = time;
f[i] = f[u]+1;
Q.removeFirst();
if(Q.size()==0)
return;
else
int v = Q.getFirst();
bfs1(v, M, process, Q, d, f);
public static void bfs_init2(ArrayList<Integer> al[], int[] process, ArrayDeque<Integer> Q, int[] d, int[] f)
Q.addLast(0精进DFS和BFS，按照流程来，秒杀