Train your English by the way:
Because of the recent rain, farmer John's farm has many puddles (represented by a NM matrix, | representative, rectangle matrix). Each area is represented by water (w) or land (.), farmer jhon wants to point out how many puddles have been formed in his rural area. A puddle is formed by a water and eight adjacent areas connected together.
Give a schematic diagram of a farm and guess how many water pits there are (square)
Input:
Two numbers N and M in the first row
Second line M characters one line
Output:
Number
Baidu Translate
Due to the recent rainfall, water has gathered in all parts of Farmer John's farm, represented by a rectangle of n x m (1 < = n < = 100; 1 < = m < = 100). Each square has water ("W") or dry land ("W"). Farmer John wants to know how many ponds have formed in his field. A pond is a set of connected squares, in which there is water, and a square is considered to be adjacent to eight neighbors. Give a chart of Farmer John's field to determine how many ponds he has.
Solution 1:
import java.util.Scanner;
public class Main {
static char[][] c;
static boolean flag[][];
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int N = sc.nextInt();
int M = sc.nextInt();
c = new char[N][M]; //assignment
flag = new boolean[c.length][c[0].length]; //assignment
for (int i = 0; i < N; i++) {
String s = sc.next();
c[i] = s.toCharArray(); //assignment
for (int j = 0; j < M; j++) {
flag[i][j] = false; //Initialization
}
}
int s = 0;
for (int i = 0; i < c.length; i++) {
for (int j = 0; j < c[0].length; j++) {
if (flag[i][j] != true) { //If this value has not been read
// flag[i][j] = true; redundant because the following functions will be assigned again. If assigned here, each value will fail to pass the / / assigned read status.
if (c[i][j] == 'W') { //Depth traversal if w
dfs(c, i, j, flag); //All conditions met
s++;
}
}
}
}
System.out.println(s);
}
}
public static void dfs(char[][] arr, int x, int y, boolean[][] temp) {
if (x < 0 || x >= c.length || y >= c[0].length || y < 0) //Jump out of range
return;
if (temp[x][y] == true) // Description visited
return;
if (arr[x][y] == '.') { //Is. That is, the condition is not met
temp[x][y] = true; //And give read status
return;
}
temp[x][y] = true; //All that's left are unread w
dfs(arr, x, y + 1, temp); // Value true for traversal of surrounding qualified positions
dfs(arr, x, y - 1, temp);
dfs(arr, x + 1, y, temp);
dfs(arr, x + 1, y + 1, temp);
dfs(arr, x + 1, y - 1, temp);
dfs(arr, x - 1, y, temp);
dfs(arr, x - 1, y + 1, temp);
dfs(arr, x - 1, y - 1, temp);
}
}Solution two:
Change W directly to.!! To indicate read
import java.util.Scanner;
public class Main {
static char[][] field;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
int N = sc.nextInt();
int M = sc.nextInt();
field = new char[N][M]; // assignment
for (int i = 0; i < N; i++) {
String s = sc.next();
field[i] = s.toCharArray(); // assignment
}
int s = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (field[i][j] == 'W') {
dfs(i, j, N, M);
s++;
}
}
}
System.out.println(s);
}
}
public static void dfs(int x, int y, int N, int M) {
field[x][y] = '.';
for (int i = -1; i < 2; i++) {
for (int j = -1; j < 2; j++) {
int nx = x + i;
int ny = y + j;
if (0 <= nx && nx < N && 0 <= ny && ny < M && field[nx][ny] == 'W')
dfs(nx, ny, N, M);
}
}
}
}