2024-01-30(树与图的深度优先遍历、广度优先遍历、拓扑排序）-CSDN博客 

2024-01-31（最短路径）-CSDN博客 

多源BFS

173. 矩阵距离 - AcWing题库

求出每个位置到所有为1的点的最短距离 

import java.util.*;
import java.io.*;

class PII{
    int x, y;
    public PII(int x, int y){
        this.x = x;
        this.y = y;
    }
}

public class Main{
    static int N = 1010, M = N * N;
    static int n, m;
    static char[][] g = new char[N][N];
    static int[][] dist = new int[N][N];
    static PII[] q = new PII[M];
    
    public static void bfs(){
        int hh = 0, tt = -1;
        
        for(int i = 0; i < N; i ++){
            Arrays.fill(dist[i], -1);//标记为没走过
        }
        
        for(int i = 0; i < n; i ++){
            for(int j = 0; j < m; j ++){
                if(g[i][j] == '1'){
                    dist[i][j] = 0;//所有为1的点都置为0
                    q[++ tt] = new PII(i, j);//把所有起点都加入队列
                }
            }
        }
        
        int[] dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};//四个方向
        
        while(hh <= tt){
            PII t = q[hh ++];
            for(int i = 0; i < 4; i ++){
                int a = t.x + dx[i];
                int b = t.y + dy[i];
                if(a < 0 || b < 0 || a >= n || b >= m) continue;//边界
                if(dist[a][b] != -1) continue;//如果走过了
                
                dist[a][b] = dist[t.x][t.y] + 1;//距离加上1
                q[++ tt] = new PII(a, b);
            }
        }
    }
    
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        String[] s = br.readLine().split(" ");
        n = Integer.parseInt(s[0]);
        m = Integer.parseInt(s[1]);
        
        for(int i = 0; i < n; i ++){//输入数据
            String str = br.readLine();
            char[] arr = str.toCharArray();
            for(int j = 0; j < m; j ++){
                g[i][j] = arr[j];
            }
        }
        
        bfs();
        
        for(int i = 0; i < n; i ++){
            for(int j = 0; j < m; j ++){
                bw.write(dist[i][j] + " ");
            }
            bw.write("\n");
        }
        bw.flush();//关闭流
    }
}

最小步数模型 

1107. 魔板 - AcWing题库 

import java.util.*;

class PII{
    char opera;
    String str;
    public PII(char opera, String str){
        this.opera = opera;
        this.str = str;
    }
}

public class Main{
    static char[][] g = new char[2][4];
    static Map<String, Integer> dist = new HashMap<>();//到达某个状态要走的距离
    static Map<String, PII> pre = new HashMap<>();//某个状态的由哪个状态经过abc哪个步骤来的
    static Queue<String> q = new LinkedList<>();//队列
    
    //将字符串转化为二维
    public static void set(String t){
        for(int i = 0; i < 4; i ++) g[0][i] = t.charAt(i);
        for(int i = 0; i < 4; i ++) g[1][i] = t.charAt(7 - i);
    }
    
    //将二维转化为字符串
    public static String get(){
        String res = "";
        for(int i = 0; i < 4; i ++) res += g[0][i];
        for(int i = 3; i >= 0; i --) res += g[1][i];
        return res;
    }
    
    //交换上下两行
    public static String move1(String t){
        set(t);
        for(int i = 0; i < 4; i ++){
            char temp = g[0][i];
            g[0][i] = g[1][i];
            g[1][i] = temp;
        }
        return get();
    }
    
    //最右边一列插入到最左边
    public static String move2(String t){
        set(t);
        char a = g[0][3];
        char b = g[1][3];
        for(int i = 3; i >= 1; i --){
            g[0][i] = g[0][i - 1];
            g[1][i] = g[1][i - 1];
        }
        g[0][0] = a;
        g[1][0] = b;
        return get();
    }
    
    //顺时针旋转
    public static String move3(String t){
        set(t);
        char temp = g[0][1];
        g[0][1] = g[1][1];
        g[1][1] = g[1][2];
        g[1][2] = g[0][2];
        g[0][2] = temp;
        return get();
    }
    
    public static int bfs(String start, String end){
        if(start.equals(end)) return 0;
        
        q.offer(start);
        dist.put(start, 0);
        
        while(!q.isEmpty()){
            String t = q.poll();
            
            String[] m = new String[3];
            m[0] = move1(t);//abc三个步骤
            m[1] = move2(t);
            m[2] = move3(t);
            
            for(int i = 0; i < 3; i ++){
                if(dist.containsKey(m[i])) continue;
                dist.put(m[i], dist.get(t) + 1);
                
                if(pre.containsKey(m[i])) continue;
                pre.put(m[i], new PII((char)('A' + i), t));
                
                q.offer(m[i]);
                
                if(m[i].equals(end)) return dist.get(end);
            }
        }
        return -1;
    }
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String start = "12345678";
        String end = "";
        for(int i = 0; i < 8; i ++){
            end += sc.next();
        }
        
        System.out.println(bfs(start, end));//步骤数
        
        String res = "";//步骤答案
        String ans = end;
        while(!ans.equals(start)){//由end倒推回start
            res += pre.get(ans).opera;
            ans = pre.get(ans).str;
        }
        
        for(int i = res.length() - 1; i >= 0; i --){
            System.out.print(res.charAt(i));
        }
    }
}

 

双端队列广搜

175. 电路维修 - AcWing题库 

这里由于起点（0，0）的横纵坐标是偶数，因为走的是斜线，所以到达的点也一定是偶数。 

import java.util.*;

class PII{
    int x, y;
    public PII(int x, int y){
        this.x = x;
        this.y = y;
    }
}

public class Main{
    static int N = 510, n, m, INF = 0x3f3f3f3f;
    static char[][] g = new char[N][N];
    static int[][] dist = new int[N][N];
    static boolean[][] st = new boolean[N][N];
    
    static int[] dx = {-1, -1, 1, 1}, dy = {-1, 1, 1, -1};//四个行进方向
    static int[] ix = {-1, -1, 0, 0}, iy = {-1, 0, 0, -1};//行进方向上面的斜杠的坐标
    static char[] s = {'\\', '/', '\\', '/'};
    
    public static int bfs(){
        //由于有多组测试数据，所以每次都要初始化
        for(int i = 0; i < N; i ++){
            Arrays.fill(st[i], false);
            Arrays.fill(dist[i], INF);
        }
        Deque<PII> q = new LinkedList<>();//双端队列
        
        dist[0][0] = 0;
        q.addLast(new PII(0, 0));
        
        while(!q.isEmpty()){
            PII t = q.pollFirst();
            int x = t.x;
            int y = t.y;
            
            if(x == n && y == m) return dist[x][y];//等于终点的时候，返回距离
            
            if(st[x][y]) continue;//遍历过了
            st[x][y] = true;
            
            for(int i = 0; i < 4; i ++){
                int a = x + dx[i];
                int b = y + dy[i];
                if(a < 0 || b < 0 || a > n || b > m) continue;//越界
                
                int ga = x + ix[i];
                int gb = y + iy[i];
                int w = (g[ga][gb] == s[i]) ? 0 : 1;//看一下能不能走，能走变0，不能走变1
                
                if(dist[a][b] > dist[x][y] + w){
                    dist[a][b] = dist[x][y] + w;//如果下一步的距离小于这个点新点的距离，那就更新
                }
                
                if(w == 0) q.addFirst(new PII(a, b));//如果是0，放在队头
                else q.addLast(new PII(a, b));//如果不是0，放在队尾
            }
        }
        return -1;
    }
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while(T -- > 0){
            n = sc.nextInt();
            m = sc.nextInt();
            for(int i = 0; i < n; i ++){
                String str = sc.next();
                for(int j = 0; j < m; j ++){
                    g[i][j] = str.charAt(j);
                }
            }
            
            int t = bfs();
            if(t == -1) System.out.println("NO SOLUTION");
            else System.out.println(t);
        }
    }
}