java dfs

/*

* Lake Counting

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 25214 Accepted: 12701

Description

Due to recent rains, water has pooled in various places in Farmer John’s field, which is

represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square

contains either water (‘W’) or dry land (‘.’). Farmer John would like to figure out how

many ponds have formed in his field. A pond is a connected set of squares with water in

them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John’s field, determine how many ponds he has.

Input

Line 1: Two space-separated integers: N and M

Lines 2..N+1: M characters per line representing one row of Farmer John’s field. Each

character is either ‘W’ or ‘.’. The characters do not have spaces between them.

Output

Line 1: The number of ponds in Farmer John’s field.

Sample Input

10 12

W……..WW.

.WWW…..WWW

….WW…WW.

………WW.

………W..

..W……W..

.W.W…..WW.

W.W.W…..W.

.W.W……W.

..W…….W.

Sample Output

3

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right

side.

* */

注意区分Java与c++的区别：
1.Java没有全局变量的概念，与之等价的是类变量
2.Java中的scanner不能输入一个字符，可以用scanner.next()再进行适当的转换得到。
3.将字符串转为用字符数组存储，调用str.toCharArray();
import java.security.PublicKey;
import java.util.Scanner;
4.可以将全局变量，被调用函数写在一个类（class global）里然后直接调用类变量，类方法即可

import org.omg.CORBA.PUBLIC_MEMBER;

class g_Var{
public static int n,m;
public static char pools[][] = new char[110][110];
public static void dfs(int x,int y){
pools[x][y] = '.';
for(int i=-1;i<=1;i++){
for(int j=-1;j<=1;j++){
int nx=x+i,ny=y+j;
if(nx>=0&&nx<n&&ny>=0&&ny<m&&pools[nx][ny]=='W'){
dfs(nx, ny);
}
}
}
return;
}
}

public class Main{

public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int cnt;
g_Var.n = sc.nextInt();//通过类名操作类变量
g_Var.m = sc.nextInt();
String str;
/*java中没有读入单个字符的方法。但可以转换 如：String s=scanner.next(); char c=s.charAt(0)
* 关键在于如何将读入的数据进行转换为char*/
for(int i=0;i<g_Var.n;i++){
str = sc.next();
g_Var.pools[i] = str.toCharArray();//调用函数转换为char数组
}
cnt = 0;
for(int i=0;i<g_Var.n;i++){
for(int j=0;j<g_Var.m;j++){
if(g_Var.pools[i][j]=='W'){
g_Var.dfs(i, j);
cnt++;
}
}
}
System.out.println(cnt);
}
}

/**

*

* 水池数目

时间限制：3000 ms | 内存限制：65535 KB

难度：4

描述

南阳理工学院校园里有一些小河和一些湖泊，现在，我们把它们通一看成水池，假设有一张我们学校的某处的地图，这个地图上仅标识了

此处是否是水池，现在，你的任务来了，请用计算机算出该地图中共有几个水池。

输入

第一行输入一个整数N，表示共有N组测试数据

每一组数据都是先输入该地图的行数m(0

import java.security.PublicKey;
import java.util.Scanner;

import org.omg.CORBA.PUBLIC_MEMBER;

class g_Var{
public static int n,m;
public static int pools[][] = new int[110][110];
final static int dirx[]={-1,1,0,0},diry[]={0,0,-1,1};
public static void dfs(int x,int y){
pools[x][y] = 0;
for(int i=0;i<4;i++){
int nx=x+g_Var.dirx[i],ny=y+g_Var.diry[i];
if(nx>=0&&nx<g_Var.n&&ny>=0&&ny<g_Var.m&&g_Var.pools[nx][ny]==1){
dfs(nx, ny);
}
}
return;
}
}

public class Main {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
int cnt;
while(T>=1){
--T;
g_Var.n = sc.nextInt();//通过类名操作类变量
g_Var.m = sc.nextInt();
for(int i=0;i<g_Var.n;i++){
for(int j=0;j<g_Var.m;j++){
g_Var.pools[i][j] = sc.nextInt();
}
}
cnt = 0;
for(int i=0;i<g_Var.n;i++){
for(int j=0;j<g_Var.m;j++){
if(g_Var.pools[i][j]==1){
g_Var.dfs(i, j);
cnt++;
}
}
}
System.out.println(cnt);
}
}
}