宝岛探险问题

问题描述：某片海域有诸多岛屿，用0表示海洋，1-9表示陆地，现给定一个岛屿上的坐标点，求解所在岛屿的面积

思路：显然这是一个搜索算法，即只要从当前坐标点开始遍历，每遍历到一个点进行计数即可，但是要注意sum的初始值为1！！！

Input：
10 10
1 2 1 0 0 0 0 0 2 3
3 0 2 0 1 2 1 0 1 2
4 0 1 0 1 2 3 2 0 1
3 2 0 0 0 1 2 4 0 0
0 0 0 0 0 0 1 5 3 0
0 1 2 1 0 1 5 4 3 0
0 1 2 3 1 3 6 2 1 0
0 0 3 4 8 9 7 5 0 0
0 0 0 3 7 8 6 0 1 2
0 0 0 0 0 0 0 0 1 0
6 8

Output：
38

DFS

import java.util.Scanner;public class DFS {static int[][] a = new int[50][50];static int[][] book = new int[50][50];static int sum = 1;static int n, m;static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {a[i][j] = input.nextInt();}}int startX = input.nextInt();int startY = input.nextInt();book[startX][startY] = 1;dfs(startX, startY);System.out.println(sum);}public static void dfs(int x, int y) {int[][] next = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};int tx, ty;for (int i = 0; i < 4; i++) {tx = x + next[i][0];ty = y + next[i][1];if(tx < 0 || tx > n - 1 || ty < 0 || ty > n - 1) {continue;}if (a[tx][ty] > 0 && book[tx][ty] == 0) {sum ++;book[tx][ty] = 1;dfs(tx, ty);}}return;}
}

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BFS

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;class node {int x;int y;node(int x, int y) {this.x = x;this.y = y;}
}
public class BFS {static int[][] a = new int[50][50];static int[][] book = new int[50][50];static int n, m;static int sum = 1;static Queue<node> queue = new LinkedList<>();static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {a[i][j] = input.nextInt();}}int startX = input.nextInt();int startY = input.nextInt();queue.offer(new node(startX, startY));book[startX][startY] = 1;bfs();System.out.println(sum);}public static void bfs() {int[][] next = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};int tx, ty;while (!queue.isEmpty()) {for (int i = 0; i < 4; i++) {tx = queue.peek().x + next[i][0];ty = queue.peek().y + next[i][1];if (tx < 0 || tx > n - 1 || ty < 0 || ty > n - 1) {continue;}if(a[tx][ty] > 0 && book[tx][ty] == 0) {queue.offer(new node(tx, ty));sum++;book[tx][ty] = 1;}}queue.remove();}return;}
}

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拓展：漫水填充法（FloodFill）
问题： 要求解区域中总共有多少岛屿？？
思路：对每个点进深搜，对于每个大于0的点进行填充负值，负值每次-1，最后输出正的这个值，即是岛屿个数。
Input:
10 10
1 2 1 0 0 0 0 0 2 3
3 0 2 0 1 2 1 0 1 2
4 0 1 0 1 2 3 2 0 1
3 2 0 0 0 1 2 4 0 0
0 0 0 0 0 0 1 5 3 0
0 1 2 1 0 1 5 4 3 0
0 1 2 3 1 3 6 2 1 0
0 0 3 4 8 9 7 5 0 0
0 0 0 3 7 8 6 0 1 2
0 0 0 0 0 0 0 0 1 0
Output:

import java.util.Scanner;public class DFS {static int[][] a = new int[50][50];static int[][] book = new int[50][50];static int sum = 1;static int num = 0;static int n, m;static Scanner input = new Scanner(System.in);public static void main(String[] args) {n = input.nextInt();m = input.nextInt();for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {a[i][j] = input.nextInt();}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (a[i][j] > 0) {num--;book[i][j] = 1;dfs(i, j, num);}}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {System.out.print(a[i][j] + "\t");}System.out.println();}System.out.println(-num);}public static void dfs(int x, int y, int color) {int[][] next = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};int tx, ty;a[x][y] = color;for (int i = 0; i < 4; i++) {tx = x + next[i][0];ty = y + next[i][1];if(tx < 0 || tx > n - 1 || ty < 0 || ty > n - 1) {continue;}if (a[tx][ty] > 0 && book[tx][ty] == 0) {sum ++;book[tx][ty] = 1;dfs(tx, ty, color);}}return;}
}