给定一个二元矩阵mat[][] ，任务是找到所有可能的非空连接单元的大小。
An empty cell is denoted by 0 while a non-empty cell is denoted by 1.
Two cells are said to be connected if they are adjacent to each other horizontally or vertically, i.e. mat[i][j] = mat[i][j – 1] or mat[i][j] = mat[i][j + 1] or mat[i][j] = mat[i – 1][j] or mat[i][j] = mat[i + 1][j].
例子：
Input: mat[][] = {{1, 1, 0, 0, 0}, {0, 1, 0, 0, 1}, {1, 0, 0, 1, 1}, {0, 0, 0, 0, 0}, {1, 0, 1, 0, 1}}
Output: 3 3 1 1 1 1
Explanation:
{mat[0][0], mat[0][1], mat[1][1]}, {mat[1][4], mat[2][3], mat[2][4]}}, {mat[2][0]}, {mat[4][0]}, {mat[4][2]}, {mat[4[4]} are the only possible connections.
Input:mat[][] = {{1, 1, 0, 0, 0}, {1, 1, 0, 1, 1}, {1, 0, 0, 1, 1}, {0, 0, 0, 0, 0}, {0, 0, 1, 1, 1}}
Output: 5 4 3
方法：
这个想法是在矩阵上使用 BFS 和递归。
请按照以下步骤操作：
- 初始化队列数据结构并插入一个单元格( mat[i][j] = 1 )。
- 对插入的单元格执行BFS并遍历其相邻单元格。
- 检查边界条件并检查当前元素是否为1 ，然后将其翻转为0 。
- 标记访问过的单元格并更新连接的非空单元格的大小。
- 最后，打印获得的连接的所有大小。
下面是上述方法的实现：
C++14
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to find size of all the
// islands from the given matrix
int BFS(vector >& mat,
int row, int col)
{
int area = 0;
// Initialize a queue for
// the BFS traversal
queue > Q;
Q.push({ row, col });
// Iterate untill the
// queue is empty
while (!Q.empty()) {
// Top element of queue
auto it = Q.front();
// Pop the element
Q.pop();
int r = it.first, c = it.second;
// Check for boundaries
if (r < 0 || c < 0 || r > 4 || c > 4)
continue;
// Check if current element is 0
if (mat[r] == 0)
continue;
// Check if current element is 1
if (mat[r] == 1) {
// Mark the cell visited
mat[r] = 0;
// Incrementing the size
area++;
}
// Traverse all neighbors
Q.push({ r + 1, c });
Q.push({ r - 1, c });
Q.push({ r, c + 1 });
Q.push({ r, c - 1 });
}
// Return the answer
return area;
}
// Function to print size of each connections
void sizeOfConnections(vector > mat)
{
// Stores the size of each
// connected non-empty
vector result;
for (int row = 0; row < 5; ++row) {
for (int col = 0; col < 5; ++col) {
// Check if the cell is
// non-empty
if (mat[row][col] == 1) {
// Function call
int area = BFS(mat, row, col);
result.push_back(area);
}
}
}
// Print the answer
for (int val : result)
cout << val << " ";
}
// Driver Code
int main()
{
vector > mat
= { { 1, 1, 0, 0, 0 },
{ 1, 1, 0, 1, 1 },
{ 1, 0, 0, 1, 1 },
{ 1, 0, 0, 0, 0 },
{ 0, 0, 1, 1, 1 } };
sizeOfConnections(mat);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
class GFG{
static class pair
{
int first, second;
pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find size of all the
// islands from the given matrix
static int BFS(int[][] mat,
int row, int col)
{
int area = 0;
// Initialize a queue for
// the BFS traversal
Queue Q = new LinkedList<>();
Q.add(new pair(row, col));
// Iterate untill the
// queue is empty
while (!Q.isEmpty())
{
// Top element of queue
pair it = Q.peek();
// Pop the element
Q.poll();
int r = it.first, c = it.second;
// Check for boundaries
if (r < 0 || c < 0 ||
r > 4 || c > 4)
continue;
// Check if current element is 0
if (mat[r] == 0)
continue;
// Check if current element is 1
if (mat[r] == 1)
{
// Mark the cell visited
mat[r] = 0;
// Incrementing the size
area++;
}
// Traverse all neighbors
Q.add(new pair(r + 1, c));
Q.add(new pair(r - 1, c));
Q.add(new pair(r, c + 1));
Q.add(new pair(r, c - 1));
}
// Return the answer
return area;
}
// Function to print size of each connections
static void sizeOfConnections(int[][] mat)
{
// Stores the size of each
// connected non-empty
ArrayList result = new ArrayList<>();
for(int row = 0; row < 5; ++row)
{
for(int col = 0; col < 5; ++col)
{
// Check if the cell is
// non-empty
if (mat[row][col] == 1)
{
// Function call
int area = BFS(mat, row, col);
result.add(area);
}
}
}
// Print the answer
for(int val : result)
System.out.print(val + " ");
}
// Driver code
public static void main (String[] args)
{
int[][] mat = { { 1, 1, 0, 0, 0 },
{ 1, 1, 0, 1, 1 },
{ 1, 0, 0, 1, 1 },
{ 1, 0, 0, 0, 0 },
{ 0, 0, 1, 1, 1 } };
sizeOfConnections(mat);
}
}
// This code is contributed by offbeat Python3
# Python3 program to implement
# the above approach
from collections import deque
# Function to find size of all the
# islands from the given matrix
def BFS(mat, row, col):
area = 0
# Initialize a queue for
# the BFS traversal
Q = deque()
Q.append([row, col])
# Iterate untill the
# queue is empty
while (len(Q) > 0):
# Top element of queue
it = Q.popleft()
# Pop the element
# Q.pop();
r, c = it[0], it[1]
# Check for boundaries
if (r < 0 or c < 0 or r > 4 or c > 4):
continue
# Check if current element is 0
if (mat[r] == 0):
continue
# Check if current element is 1
if (mat[r] == 1):
# Mark the cell visited
mat[r] = 0
# Incrementing the size
area += 1
# Traverse all neighbors
Q.append([r + 1, c])
Q.append([r - 1, c])
Q.append([r, c + 1])
Q.append([r, c - 1])
# Return the answer
return area
# Function to prsize of each connections
def sizeOfConnections(mat):
# Stores the size of each
# connected non-empty
result = []
for row in range(5):
for col in range(5):
# Check if the cell is
# non-empty
if (mat[row][col] == 1):
# Function call
area = BFS(mat, row, col);
result.append(area)
# Print the answer
for val in result:
print(val, end = " ")
# Driver Code
if __name__ == '__main__':
mat = [ [ 1, 1, 0, 0, 0 ],
[ 1, 1, 0, 1, 1 ],
[ 1, 0, 0, 1, 1 ],
[ 1, 0, 0, 0, 0 ],
[ 0, 0, 1, 1, 1 ] ]
sizeOfConnections(mat)
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find size of all the
// islands from the given matrix
static int BFS(int[, ] mat, int row, int col)
{
int area = 0;
// Initialize a queue for
// the BFS traversal
Queue Q = new Queue();
Q.Enqueue(new pair(row, col));
// Iterate untill the
// queue is empty
while (Q.Count != 0)
{
// Top element of queue
pair it = Q.Peek();
// Pop the element
Q.Dequeue();
int r = it.first, c = it.second;
// Check for boundaries
if (r < 0 || c < 0 || r > 4 || c > 4)
continue;
// Check if current element is 0
if (mat[r, c] == 0)
continue;
// Check if current element is 1
if (mat[r, c] == 1)
{
// Mark the cell visited
mat[r, c] = 0;
// Incrementing the size
area++;
}
// Traverse all neighbors
Q.Enqueue(new pair(r + 1, c));
Q.Enqueue(new pair(r - 1, c));
Q.Enqueue(new pair(r, c + 1));
Q.Enqueue(new pair(r, c - 1));
}
// Return the answer
return area;
}
// Function to print size of each connections
static void sizeOfConnections(int[,] mat)
{
// Stores the size of each
// connected non-empty
ArrayList result = new ArrayList();
for(int row = 0; row < 5; ++row)
{
for(int col = 0; col < 5; ++col)
{
// Check if the cell is
// non-empty
if (mat[row, col] == 1)
{
// Function call
int area = BFS(mat, row, col);
result.Add(area);
}
}
}
// Print the answer
foreach(int val in result)
Console.Write(val + " ");
}
// Driver code
public static void Main(string[] args)
{
int[, ] mat = { { 1, 1, 0, 0, 0 },
{ 1, 1, 0, 1, 1 },
{ 1, 0, 0, 1, 1 },
{ 1, 0, 0, 0, 0 },
{ 0, 0, 1, 1, 1 } };
sizeOfConnections(mat);
}
}
// This code is contributed by grand_master 输出：
6 4 3时间复杂度： O(row * col)
辅助空间： O(row * col)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live