给定 ‘W’和‘L’的 2D 网格arr[][] ，其中‘W’表示水， ‘L’表示陆地，任务是找到必须更改为陆地的水分量 ‘W’ 的最小数量组件“L”，使两个岛连接起来。
An island is the set of connected ‘L’s.
注意：只能有两个不相交的岛。
例子：
Input: arr[][] = {{‘W’, ‘L’}, {‘L’, ‘W’}};
Output: 1
Explanation:
For the given set of islands if we change arr[1][1] to ‘W’ then, set of all island are connected.
Therefore, the minimum number of ‘W’ must be changed to ‘L’ is 1.
Input: arr[][] = {{‘W’, ‘L’, ‘W’}, {‘W’, ‘W’, ‘W’}, {‘W’, ‘W’, ‘L’}}
Output: 2
方法：这个问题可以使用 Floodfill 算法解决。以下是步骤：
- 对第一组连接的岛屿使用 Floodfill 算法，并使所有岛屿都被访问并将坐标存储在哈希中（比如hash1 ）。
- 对第二组连接的岛屿使用 Floodfill 算法，并使所有岛屿都被访问，并将坐标存储在第二个哈希中（比如hash2 ）。
- 存储在散列中的坐标之间的最小差异就是所需的结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Determine the distance between two
// coordinates
int dist(pair& p1,
pair& p2)
{
return abs(p1.first - p2.first)
+ abs(p2.second - p1.second) - 1;
}
// Function to implement floodfill algorithm
void floodfill(set >& hash,
int i, int j,
vector >& A)
{
// Base Case
if (i < 0 || i >= A.size() || j < 0
|| j >= A[0].size() || A[i][j] != 'L') {
return;
}
// Mark the current node visited
A[i][j] = 'V';
// Store the coordinates of in the
// hash set
hash.insert(make_pair(i, j));
// Recursively iterate for the next
// four directions
floodfill(hash, i - 1, j, A);
floodfill(hash, i + 1, j, A);
floodfill(hash, i, j - 1, A);
floodfill(hash, i, j + 1, A);
}
// Function to find the minimum 'W' to flipped
void findMinimumW(vector >& A)
{
// Base Case
if (A.size() == 0)
return;
// Two sets to store the coordinates of
// connected island
set > hash1, hash2;
// Traversing the given grid[][]
for (int i = 0; i < A.size(); i++) {
for (int j = 0; j < A[0].size(); j++) {
// If an island is found
if (A[i][j] == 'L') {
// Floodfill Algorithm for
// the first island
if (hash1.empty()) {
floodfill(hash1, i, j, A);
}
// Floodfill Algorithm for
// the second island
if (hash2.empty()
&& !hash1.count({ i, j })) {
floodfill(hash2, i, j, A);
}
}
}
}
// To store the minimum count of 'W'
int ans = INT_MAX;
// Traverse both pairs of hashes
for (auto i : hash1) {
for (auto j : hash2) {
ans = min(ans, dist(i, j));
}
}
// Print the final answer
cout << ans << endl;
}
// Driver Code
int main()
{
// Given grid of land and water
vector > arr;
arr = { { 'W', 'L' }, { 'L', 'W' } };
// Function Call
findMinimumW(arr);
return 0;
} Python3
# Python3 program for the above approach
import sys
# Determine the distance between two
# coordinates
def dist(p1, p2):
return (abs(p1[0] - p2[0]) +
abs(p2[1] - p1[1]) - 1)
# Function to implement floodfill algorithm
def floodfill(hash, i, j, A):
# Base Case
if (i < 0 or i >= len(A) or j < 0 or
j >= len(A[0]) or A[i][j] != 'L'):
return hash, A
# Mark the current node visited
A[i][j] = 'V'
# Store the coordinates of in the
# hash set
hash.add((i, j))
# Recursively iterate for the next
# four directions
hash, A = floodfill(hash, i - 1, j, A)
hash, A = floodfill(hash, i + 1, j, A)
hash, A = floodfill(hash, i, j - 1, A)
hash, A = floodfill(hash, i, j + 1, A)
return hash, A
# Function to find the minimum 'W' to flipped
def findMinimumW(A):
# Base Case
if (len(A) == 0):
return set(), A
# Two sets to store the coordinates of
# connected island
hash1 = set()
hash2 = set()
# Traversing the given grid[][]
for i in range(len(A)):
for j in range(len(A[0])):
# If an island is found
if (A[i][j] == 'L'):
# Floodfill Algorithm for
# the first island
if (len(hash1) == 0):
hash1, A = floodfill(hash1, i, j, A)
# Floodfill Algorithm for
# the second island
if (len(hash2) == 0 and
(i, j) not in hash1):
hash2, A = floodfill(hash2, i, j, A)
# To store the minimum count of 'W'
ans = sys.maxsize
# Traverse both pairs of hashes
for i in hash1:
for j in hash2:
ans = min(ans, dist(i, j))
# Print the final answer
print(ans)
# Driver Code
if __name__=='__main__':
# Given grid of land and water
arr = []
arr = [ [ 'W', 'L' ], [ 'L', 'W' ] ]
# Function Call
findMinimumW(arr)
# This code is contributed by pratham76输出：
1时间复杂度： O(N 2 )
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。