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代码随想录打卡|Day45 图论（孤岛的总面积、沉没孤岛、水流问题、建造最大岛屿）

news 来源：原创 2025/5/22 7:10:50

图论part03

孤岛的总面积

代码随想录链接
题目链接
视频讲解链接

在这里插入图片描述

思路：既然某个网格在边界上的岛屿不是孤岛，那么就把非孤岛的所有岛屿变成海洋，最后再次统计还剩余的岛屿占据的网格总数即可。

dfs：

import java.util.Scanner;public class Main{static int res = 0;static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};// TODO 4.dfs遍历逻辑private static void dfs(int[][] graph , int x , int y){graph[x][y] = 0;for(int i = 0 ; i < 4 ; i++){int nextX = x + dir[i][0];int nextY = y + dir[i][1];if(nextX < 0 || nextY < 0 || nextX >= graph.length || nextY >= graph[0].length) continue;if(graph[nextX][nextY] == 1){graph[nextX][nextY] = 0;dfs(graph,nextX,nextY);}}}public static void main(String[] args){// TODO 1.生成graphScanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int[][] graph = new int[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){graph[i][j] = sc.nextInt();}}// sc.colse();// TODO 2.仅对四个边进行遍历，遇到陆地的时候将该陆地和与之相邻的陆地全部变为海洋for(int i = 0; i < n ; i++){if(graph[i][0] == 1){dfs(graph,i,0);}}for(int i = 0; i < n ; i++){if(graph[i][m - 1] == 1){dfs(graph,i,m-1);}}for(int j = 0 ; j < m ; j++){if(graph[0][j] == 1){dfs(graph,0,j);}}for(int j = 0 ; j < m ; j++){if(graph[n - 1][j] == 1){dfs(graph,n-1,j);}}// TODO 3.遍历图，获取所有岛屿土地数量int sum = 0 ;for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(graph[i][j] == 1 ){sum++;                }}}System.out.println(sum);}
}

bfs：

import java.util.*;public class Main{static int res = 0;static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};// TODO 4.bfs遍历逻辑private static void bfs(int[][] graph , int x , int y){class Node{int x;int y;public Node(int x , int y){this.x = x;this.y = y;}}Queue<Node> queue = new LinkedList<>();queue.add(new Node(x,y));graph[x][y] = 0;while(!queue.isEmpty()){Node node = queue.remove();for(int i = 0 ; i < 4 ; i++){int nextX = node.x + dir[i][0];int nextY = node.y + dir[i][1];if(nextX < 0 || nextY < 0 || nextX >= graph.length || nextY >= graph[0].length)continue;if(graph[nextX][nextY] == 1){graph[nextX][nextY] = 0;queue.add(new Node(nextX,nextY));}}}}public static void main(String[] args){// TODO 1.生成graphScanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int[][] graph = new int[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){graph[i][j] = sc.nextInt();}}// sc.colse();// TODO 2.仅对四个边进行遍历，遇到陆地的时候将该陆地和与之相邻的陆地全部变为海洋for(int i = 0; i < n ; i++){if(graph[i][0] == 1){bfs(graph,i,0);}}for(int i = 0; i < n ; i++){if(graph[i][m - 1] == 1){bfs(graph,i,m-1);}}for(int j = 0 ; j < m ; j++){if(graph[0][j] == 1){bfs(graph,0,j);}}for(int j = 0 ; j < m ; j++){if(graph[n - 1][j] == 1){bfs(graph,n-1,j);}}// TODO 3.遍历图，获取所有岛屿土地数量int sum = 0 ;for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(graph[i][j] == 1 ){sum++;                }}}System.out.println(sum);}
}

沉没孤岛

代码随想录链接
题目链接
视频讲解链接

在这里插入图片描述

思路：在这道题之中，我们只需要将图保留两份，一份计算孤岛，计算完之后对第二份图的孤岛进行去除即可。

DFS:

import java.util.Scanner;public class Main{static int res = 0;static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};// TODO 4.dfs遍历逻辑private static void dfs(int[][] graph , int x , int y){graph[x][y] = 0;for(int i = 0 ; i < 4 ; i++){int nextX = x + dir[i][0];int nextY = y + dir[i][1];if(nextX < 0 || nextY < 0 || nextX >= graph.length || nextY >= graph[0].length) continue;if(graph[nextX][nextY] == 1){graph[nextX][nextY] = 0;dfs(graph,nextX,nextY);}}}public static void main(String[] args){// TODO 1.生成graphScanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int[][] graph = new int[n][m];int[][] graph2 = new int[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){int num = sc.nextInt();graph[i][j] = num;graph2[i][j] = num;}}// sc.colse();// TODO 2.仅对四个边进行遍历，遇到陆地的时候将该陆地和与之相邻的陆地全部变为海洋for(int i = 0; i < n ; i++){if(graph[i][0] == 1){dfs(graph,i,0);}}for(int i = 0; i < n ; i++){if(graph[i][m - 1] == 1){dfs(graph,i,m-1);}}for(int j = 0 ; j < m ; j++){if(graph[0][j] == 1){dfs(graph,0,j);}}for(int j = 0 ; j < m ; j++){if(graph[n - 1][j] == 1){dfs(graph,n-1,j);}}// TODO 3.遍历图，获取所有岛屿土地数量for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(graph[i][j] == 1 && graph2[i][j] == 1){graph2[i][j] = 0;                }System.out.print(graph2[i][j] + " ");}System.out.println();}// for(int i = 0 ; i < n ; i++){//     for(int j = 0 ; j < m ; j++){//         System.out.print(graph2[i][j] + " ");//     }//     System.out.println();// }}
}

BFS：

import java.util.*;public class Main{static int res = 0;static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};// TODO 4.bfs遍历逻辑private static void bfs(int[][] graph , int x , int y){class Node{int x;int y;public Node(int x , int y){this.x = x;this.y = y;}}Queue<Node> queue = new LinkedList<>();queue.add(new Node(x,y));graph[x][y] = 0;while(!queue.isEmpty()){Node node = queue.remove();for(int i = 0 ; i < 4 ; i++){int nextX = node.x + dir[i][0];int nextY = node.y + dir[i][1];if(nextX < 0 || nextY < 0 || nextX >= graph.length || nextY >= graph[0].length)continue;if(graph[nextX][nextY] == 1){graph[nextX][nextY] = 0;queue.add(new Node(nextX,nextY));}}}}public static void main(String[] args){// TODO 1.生成graphScanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int[][] graph = new int[n][m];int[][] graph2 = new int[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){int num = sc.nextInt();graph[i][j] = num;graph2[i][j] = num;}}// sc.colse();// TODO 2.仅对四个边进行遍历，遇到陆地的时候将该陆地和与之相邻的陆地全部变为海洋for(int i = 0; i < n ; i++){if(graph[i][0] == 1){bfs(graph,i,0);}}for(int i = 0; i < n ; i++){if(graph[i][m - 1] == 1){bfs(graph,i,m-1);}}for(int j = 0 ; j < m ; j++){if(graph[0][j] == 1){bfs(graph,0,j);}}for(int j = 0 ; j < m ; j++){if(graph[n - 1][j] == 1){bfs(graph,n-1,j);}}// TODO 3.遍历图，获取所有岛屿土地数量for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(graph[i][j] == 1  && graph2[i][j] == 1){graph2[i][j] = 0;                }System.out.print(graph2[i][j] + " ");}System.out.println();}}
}

水流问题

代码随想录链接
题目链接
视频讲解链接
在这里插入图片描述

暴力DFS：

import java.util.*;public class Main{static int m,n;// static int[][] dir = {{0,1},{1,0},{0,-1},{1-,0}};static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};private static void dfs(int[][] graph , boolean[][] visted , int x , int y ){if(visted[x][y]) return ;visted[x][y] = true;for(int i = 0 ; i < 4 ; i++){int nextX = x + dir[i][0];int nextY = y + dir[i][1];if(nextX < 0 || nextY < 0 || nextX >= n || nextY >= m) continue;if(graph[x][y] < graph[nextX][nextY]) continue;dfs(graph,visted,nextX,nextY);}}private static boolean isResult(int[][] graph , int x , int y){boolean[][] visted = new boolean[n][m];dfs(graph,visted,x,y);boolean isFirst = false;boolean isSecond = false;// 判断当前的（x,y）节点是否从第一组边界出去for(int i = 0 ; i < n ; i++){if(visted[i][0]){isFirst = true;break;}}for(int j = 0 ; j < m ; j++){if(visted[0][j]){isFirst = true;break;}}// 判断当前的（x,y）节点是否从第二组边界出去for(int i = 0 ; i < n ; i++){if(visted[i][m-1]){isSecond = true;break;}}for(int j = 0 ; j < m ; j++){if(visted[n - 1][j]){isSecond = true;break;}}return isFirst&&isSecond;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();int[][] graph = new int[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){graph[i][j] = sc.nextInt();}} for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(isResult(graph,i,j))System.out.println(i + " " + j);}}}
}

暴力BFS：(超时)

import java.util.*;public class Main{static int m,n;// static int[][] dir = {{0,1},{1,0},{0,-1},{1-,0}};static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};private static void bfs(int[][] graph , boolean[][] visted , int x , int y ){class Node{int x ;int y;public Node(int x , int y){this.x = x;this.y = y;}}Queue<Node> queue = new LinkedList<>();queue.add(new Node(x,y));visted[x][y] = true;while(!queue.isEmpty()){Node node = queue.remove();for(int i = 0 ; i < 4 ; i++){int nextX = node.x + dir[i][0];int nextY = node.y + dir[i][1];if(nextX < 0 || nextY < 0 || nextX >= n || nextY >= m ) continue;if(visted[nextX][nextY]) continue;if(graph[node.x][node.y] < graph[nextX][nextY] ) continue;queue.add(new Node(nextX,nextY));visted[nextX][nextY] = true;}}}private static boolean isResult(int[][] graph , int x , int y){boolean[][] visted = new boolean[n][m];bfs(graph,visted,x,y);boolean isFirst = false;boolean isSecond = false;// 判断当前的（x,y）节点是否从第一组边界出去for(int i = 0 ; i < n ; i++){if(visted[i][0]){isFirst = true;break;}}for(int j = 0 ; j < m ; j++){if(visted[0][j]){isFirst = true;break;}}// 判断当前的（x,y）节点是否从第二组边界出去for(int i = 0 ; i < n ; i++){if(visted[i][m-1]){isSecond = true;break;}}for(int j = 0 ; j < m ; j++){if(visted[n - 1][j]){isSecond = true;break;}}return isFirst&&isSecond;}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();int[][] graph = new int[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){graph[i][j] = sc.nextInt();}} for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(isResult(graph,i,j))System.out.println(i + " " + j);}}}
}

DFS优化：
采用逆向思维，考虑水流从低处往高处可以流通的情况，从而获得第一组边界的流通情况和第二组流水的情况，二者均为true的情况则表明该网格可以从第一组边界和第二组边界流出。

import java.util.*;public class Main{static int m,n;// static int[][] dir = {{0,1},{1,0},{0,-1},{1-,0}};static int[][] dir = {{0,1},{1,0},{0,-1},{-1,0}};private static void dfs(int[][] graph , boolean[][] visted , int x , int y , int preH){// 遇到边界或者访问过的点直接返回if(x < 0 || y < 0 || x >= graph.length ||y >= graph[0].length || visted[x][y]) return;if(graph[x][y] < preH) return ;visted[x][y] = true;dfs(graph,visted,x+1,y,graph[x][y]);dfs(graph,visted,x,y+1,graph[x][y]);dfs(graph,visted,x-1,y,graph[x][y]);dfs(graph,visted,x,y-1,graph[x][y]);}public static void main(String[] args){Scanner sc = new Scanner(System.in);n = sc.nextInt();m = sc.nextInt();int[][] graph = new int[n][m];boolean[][] firstB = new boolean[n][m];boolean[][] secondB = new boolean[n][m];for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){graph[i][j] = sc.nextInt();}} // 从上下边界进行DFSfor(int j = 0 ; j < m ; j++){dfs(graph,firstB,0,j,Integer.MIN_VALUE);dfs(graph,secondB,n-1,j,Integer.MIN_VALUE);}// 从左右边界进行DFSfor(int i = 0 ; i < n ; i++){dfs(graph,firstB,i,0,Integer.MIN_VALUE);dfs(graph,secondB,i,m-1,Integer.MIN_VALUE);}for(int i = 0 ; i < n ; i++){for(int j = 0 ; j < m ; j++){if(firstB[i][j] && secondB[i][j])System.out.println(i + " " + j);}}}
}

BFS（待补充）

建造最大岛屿(待完成)

代码随想录链接
题目链接
视频讲解链接

在这里插入图片描述

import java.util.*;
public class Main {// 该方法采用 DFS// 定义全局变量// 记录每次每个岛屿的面积static int count;// 对每个岛屿进行标记static int mark;// 定义二维数组表示四个方位static int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};// DFS 进行搜索，将每个岛屿标记为不同的数字public static void dfs(int[][] grid, int x, int y, boolean[][] visited) {// 当遇到边界，直接returnif (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) return;// 遇到已经访问过的或者遇到海水，直接返回if (visited[x][y] || grid[x][y] == 0) return;visited[x][y] = true;count++;grid[x][y] = mark;// 继续向下层搜索dfs(grid, x, y + 1, visited);dfs(grid, x, y - 1, visited);dfs(grid, x + 1, y, visited);dfs(grid, x - 1, y, visited);}public static void main (String[] args) {// 接收输入Scanner sc = new Scanner(System.in);int m = sc.nextInt();int n = sc.nextInt();int[][] grid = new int[m][n];for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {grid[i][j] = sc.nextInt();}}// 初始化mark变量，从2开始（区别于0水，1岛屿）mark = 2;// 定义二位boolean数组记录该位置是否被访问boolean[][] visited = new boolean[m][n];// 定义一个HashMap，记录某片岛屿的标记号和面积HashMap<Integer, Integer> getSize = new HashMap<>();// 定义一个HashSet，用来判断某一位置水四周是否存在不同标记编号的岛屿HashSet<Integer> set = new HashSet<>();// 定义一个boolean变量，看看DFS之后，是否全是岛屿boolean isAllIsland = true;// 遍历二维数组进行DFS搜索，标记每片岛屿的编号，记录对应的面积for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == 0) isAllIsland = false;if (grid[i][j] == 1) {count = 0;dfs(grid, i, j, visited);getSize.put(mark, count);mark++;}}}int result = 0;if (isAllIsland) result =  m * n;// 对标记完的grid继续遍历，判断每个水位置四周是否有岛屿，并记录下四周不同相邻岛屿面积之和// 每次计算完一个水位置周围可能存在的岛屿面积之和，更新下result变量for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == 0) {set.clear();// 当前水位置变更为岛屿，所以初始化为1int curSize = 1;for (int[] dir : dirs) {int curRow = i + dir[0];int curCol = j + dir[1];if (curRow < 0 || curRow >= m || curCol < 0 || curCol >= n) continue;int curMark = grid[curRow][curCol];// 如果当前相邻的岛屿已经遍历过或者HashMap中不存在这个编号，继续搜索if (set.contains(curMark) || !getSize.containsKey(curMark)) continue;set.add(curMark);curSize += getSize.get(curMark);}result = Math.max(result, curSize);}}}// 打印结果System.out.println(result);}
}