AC code:
1. bfs
import javax.swing.*;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.nio.file.attribute.AclEntryFlag;
import java.security.AlgorithmConstraints;
import java.sql.Struct;
import java.text.CollationElementIterator;
import java.text.DateFormatSymbols;
import java.util.*;
import java.util.stream.Collectors;
public class Main
{
static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
static int N = (int)0 + 20;
static math_myself math_me = new math_myself();
static char g[][] = new char[N][N];
static boolean used[][] = new boolean[N][N];
static int dx[] = {-1,0,1,0},dy[] = {0,1,0,-1};
static Queue<PII> q = new LinkedList<>();
static int n,m;
static int bfs(int x, int y)
{
q.add(new PII(x,y));
used[x][y] = true;
int cnt = 1; // The starting point is the black brick, so it should also be counted
while(q. size() > 0)
{
PII t = q.poll();
for(int i = 0 ; i < 4 ; i ++ )
{
int x_cur = t.x + dx[i],y_cur = t.y + dy[i];
if(x_cur < 0 || x_cur >= n || y_cur < 0 || y_cur >= m) continue; // Do not go beyond the boundary
if (used[x_cur][y_cur]) continue; // The tiles that have passed will not go
if (g[x_cur][y_cur] != '.') continue; // If it is not a brand new black tile, it will not go away
// Brand new black tiles must be walkable
used[x_cur][y_cur] = true;
q.add(new PII(x_cur,y_cur));
cnt + + ;
}
}
return cnt;
}
public static void main(String[] args ) throws IOException
{
while(true)
{
q. clear();
for(int i = 0 ; i < n; i ++ ) Arrays.fill(used[i],false);
// Normally, input n first, then m, this question is reversed
m = rd.nextInt();
n = rd.nextInt();
if (m == 0 || n == 0) break;
// read the matrix
for(int i = 0 ; i < n ; i ++ ) g[i] = rd.next().toCharArray();
// find the starting point
int x = 0, y = 0, flag = 0;
for(int i = 0 ; i < n ; i ++ )
{
for(int j = 0 ; j < m ; j ++ )
{
if(g[i][j] == '@')
{
x = i;
y = j;
flag = 1;
}
if (flag == 1) break;
}
}
pw. println(bfs(x,y));
}
pw.flush();
}
}
class rd
{
static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer tokenizer = new StringTokenizer("");
static String nextLine() throws IOException { return reader. readLine(); }
static String next() throws IOException
{
while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(reader.readLine());
return tokenizer. nextToken();
}
static int nextInt() throws IOException { return Integer. parseInt(next()); }
static double nextDouble() throws IOException { return Double. parseDouble(next()); }
static long nextLong() throws IOException { return Long. parseLong(next());}
static BigInteger nextBigInteger() throws IOException
{
BigInteger d = new BigInteger(rd. nextLine());
return d;
}
}
class PII
{
int x,y;
public PII(int x ,int y)
{
this.x = x;
this.y = y;
}
}
class math_myself
{
int gcd(int a, int b)
{
if(b == 0) return a;
else return gcd(b,a % b);
}
int lcm(int a,int b)
{
return a * b / gcd(a, b);
}
// Find all divisors of n
List get_factor(int n)
{
List<Long> a = new ArrayList<>();
for(long i = 1; i <= Math. sqrt(n) ; i ++ )
{
if(n % i == 0)
{
a.add(i);
if(i != n / i) a.add(n / i); // // Avoid the situation: when x = 16, i = 4, x / i = 4, this will add two situations ^-^How much can the complexity be reduced?
}
}
// Deduplication of the same factor, this method is perfect
a = a. stream(). distinct(). collect(Collectors. toList());
// Sort the factors (ascending)
Collections. sort(a);
return a;
}
// Check if it is a prime number
boolean check_isPrime(int n)
{
if(n < 2) return false;
for(int i = 2 ; i <= n / i; i ++ ) if (n % i == 0) return false;
return true;
}
}
2. dfs
import javax.swing.*;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.nio.file.attribute.AclEntryFlag;
import java.security.AlgorithmConstraints;
import java.sql.Struct;
import java.text.CollationElementIterator;
import java.text.DateFormatSymbols;
import java.util.*;
import java.util.stream.Collectors;
public class Main
{
static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
static int N = (int)0 + 20;
static math_myself math_me = new math_myself();
static char g[][] = new char[N][N];
static int dx[] = {-1,0,1,0},dy[] = {0,1,0,-1};
static int n,m;
static int cnt;
// Only brand new black tiles will enter dfs
static void dfs(int x, int y)
{
g[x][y] = '#'; // The points that have been passed become red tiles to avoid repeated walking
cnt + + ; // the black ones that walked
for(int i = 0 ; i < 4 ; i ++ )
{
int x_cur = x + dx[i],y_cur = y + dy[i];
if(x_cur < 0 || x_cur >= n || y_cur < 0 || y_cur >= m || g[x_cur][y_cur] == '#') continue; // red tiles or black tiles walked (turns into a red tile) will not enter dfs
dfs(x_cur,y_cur);
}
}
public static void main(String[] args ) throws IOException
{
while(true)
{
cnt = 0;
// Normally, input n first, then m, this question is reversed
m = rd.nextInt();
n = rd.nextInt();
if (m == 0 || n == 0) break;
// read the matrix
for(int i = 0 ; i < n ; i ++ ) g[i] = rd.next().toCharArray();
// find the starting point
int x = 0, y = 0, flag = 0;
for(int i = 0 ; i < n ; i ++ )
{
for(int j = 0 ; j < m ; j ++ )
{
if(g[i][j] == '@')
{
x = i;
y = j;
flag = 1;
}
if (flag == 1) break;
}
}
dfs(x,y); // pass in the starting point
pw. println(cnt);
}
pw.flush();
}
}
class rd
{
static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
static StringTokenizer tokenizer = new StringTokenizer("");
static String nextLine() throws IOException { return reader. readLine(); }
static String next() throws IOException
{
while (!tokenizer.hasMoreTokens()) tokenizer = new StringTokenizer(reader.readLine());
return tokenizer. nextToken();
}
static int nextInt() throws IOException { return Integer. parseInt(next()); }
static double nextDouble() throws IOException { return Double. parseDouble(next()); }
static long nextLong() throws IOException { return Long. parseLong(next());}
static BigInteger nextBigInteger() throws IOException
{
BigInteger d = new BigInteger(rd. nextLine());
return d;
}
}
class PII
{
int x,y;
public PII(int x ,int y)
{
this.x = x;
this.y = y;
}
}
class math_myself
{
int gcd(int a, int b)
{
if(b == 0) return a;
else return gcd(b,a % b);
}
int lcm(int a,int b)
{
return a * b / gcd(a, b);
}
// Find all divisors of n
List get_factor(int n)
{
List<Long> a = new ArrayList<>();
for(long i = 1; i <= Math. sqrt(n) ; i ++ )
{
if(n % i == 0)
{
a.add(i);
if(i != n / i) a.add(n / i); // // Avoid the situation: when x = 16, i = 4, x / i = 4, this will add two situations ^-^How much can the complexity be reduced?
}
}
// Deduplication of the same factor, this method is perfect
a = a. stream(). distinct(). collect(Collectors. toList());
// Sort the factors (ascending)
Collections. sort(a);
return a;
}
// Check if it is a prime number
boolean check_isPrime(int n)
{
if(n < 2) return false;
for(int i = 2 ; i <= n / i; i ++ ) if (n % i == 0) return false;
return true;
}
}