给定一个带有4种类型的块的grid [] []网格：
- 1代表起始块。恰好有一个起点。
- 2表示结束块。恰好有一个结尾块。
- 0代表我们可以走过去的空白块。
- -1表示我们无法越过的障碍。
任务是计算从起始块到结束块的路径数，以使每个无障碍块都被准确覆盖一次。
例子：
Input: grid[][] = {
{1, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 2, -1} }
Output: 2
Following are the only paths covering all the non-obstacle blocks:
Input: grid[][] = {
{1, 0, 0, 0},
{0, 0, 0, 0},
{0, 0, 0, 2} }
Output: 4
方法：我们可以在这里使用简单的DFS进行回溯。我们可以通过计算途中遇到的所有障碍物，并最终将其与可用的障碍物总数进行比较(如果它们匹配)，来检查特定路径是否覆盖了所有非障碍障碍物，然后将其添加为有效解决方案。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function for dfs.
// i, j ==> Current cell indexes
// vis ==> To mark visited cells
// ans ==> Result
// z ==> Current count 0s visited
// z_count ==> Total 0s present
void dfs(int i, int j, vector >& grid,
vector >& vis, int& ans,
int z, int z_count)
{
int n = grid.size(), m = grid[0].size();
// Mark the block as visited
vis[i][j] = 1;
if (grid[i][j] == 0)
// update the count
z++;
// If end block reached
if (grid[i][j] == 2) {
// If path covered all the non-
// obstacle blocks
if (z == z_count)
ans++;
vis[i][j] = 0;
return;
}
// Up
if (i >= 1 && !vis[i - 1][j] && grid[i - 1][j] != -1)
dfs(i - 1, j, grid, vis, ans, z, z_count);
// Down
if (i < n - 1 && !vis[i + 1][j] && grid[i + 1][j] != -1)
dfs(i + 1, j, grid, vis, ans, z, z_count);
// Left
if (j >= 1 && !vis[i][j - 1] && grid[i][j - 1] != -1)
dfs(i, j - 1, grid, vis, ans, z, z_count);
// Right
if (j < m - 1 && !vis[i][j + 1] && grid[i][j + 1] != -1)
dfs(i, j + 1, grid, vis, ans, z, z_count);
// Unmark the block (unvisited)
vis[i][j] = 0;
}
// Function to return the count of the unique paths
int uniquePaths(vector >& grid)
{
int z_count = 0; // Total 0s present
int n = grid.size(), m = grid[0].size();
int ans = 0;
vector > vis(n, vector(m, 0));
int x, y;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
// Count non-obstacle blocks
if (grid[i][j] == 0)
z_count++;
else if (grid[i][j] == 1) {
// Starting position
x = i, y = j;
}
}
}
dfs(x, y, grid, vis, ans, 0, z_count);
return ans;
}
// Driver code
int main()
{
vector > grid{ { 1, 0, 0, 0 },
{ 0, 0, 0, 0 },
{ 0, 0, 2, -1 } };
cout << uniquePaths(grid);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
static int ans = 0;
// Function for dfs.
// i, j ==> Current cell indexes
// vis ==> To mark visited cells
// ans ==> Result
// z ==> Current count 0s visited
// z_count ==> Total 0s present
static void dfs(int i, int j, int[][] grid,
boolean[][] vis, int z, int z_count)
{
int n = grid.length, m = grid[0].length;
// Mark the block as visited
vis[i][j] = true;
if (grid[i][j] == 0)
// update the count
z++;
// If end block reached
if (grid[i][j] == 2)
{
// If path covered all the non-
// obstacle blocks
if (z == z_count)
ans++;
vis[i][j] = false;
return;
}
// Up
if (i >= 1 && !vis[i - 1][j] && grid[i - 1][j] != -1)
dfs(i - 1, j, grid, vis, z, z_count);
// Down
if (i < n - 1 && !vis[i + 1][j] && grid[i + 1][j] != -1)
dfs(i + 1, j, grid, vis, z, z_count);
// Left
if (j >= 1 && !vis[i][j - 1] && grid[i][j - 1] != -1)
dfs(i, j - 1, grid, vis, z, z_count);
// Right
if (j < m - 1 && !vis[i][j + 1] && grid[i][j + 1] != -1)
dfs(i, j + 1, grid, vis, z, z_count);
// Unmark the block (unvisited)
vis[i][j] = false;
}
// Function to return the count of the unique paths
static int uniquePaths(int[][] grid)
{
int z_count = 0; // Total 0s present
int n = grid.length, m = grid[0].length;
boolean[][] vis = new boolean[n][m];
for (int i = 0; i < n; i++)
{
Arrays.fill(vis[i], false);
}
int x = 0, y = 0;
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < m; ++j)
{
// Count non-obstacle blocks
if (grid[i][j] == 0)
z_count++;
else if (grid[i][j] == 1)
{
// Starting position
x = i;
y = j;
}
}
}
dfs(x, y, grid, vis, 0, z_count);
return ans;
}
// Driver code
public static void main(String[] args)
{
int[][] grid = { { 1, 0, 0, 0 }, { 0, 0, 0, 0 }, { 0, 0, 2, -1 } };
System.out.println(uniquePaths(grid));
}
}
// This code is contributed by sanjeev2552Python3
# Python3 implementation of the approach
# Function for dfs.
# i, j ==> Current cell indexes
# vis ==> To mark visited cells
# ans ==> Result
# z ==> Current count 0s visited
# z_count ==> Total 0s present
def dfs(i, j, grid, vis, ans, z, z_count):
n = len(grid)
m = len(grid[0])
# Mark the block as visited
vis[i][j] = 1
if (grid[i][j] == 0):
# Update the count
z += 1
# If end block reached
if (grid[i][j] == 2):
# If path covered all the non-
# obstacle blocks
if (z == z_count):
ans += 1
vis[i][j] = 0
return grid, vis, ans
# Up
if (i >= 1 and not vis[i - 1][j] and
grid[i - 1][j] != -1):
grid, vis, ans = dfs(i - 1, j, grid,
vis, ans, z,
z_count)
# Down
if (i < n - 1 and not vis[i + 1][j] and
grid[i + 1][j] != -1):
grid, vis, ans = dfs(i + 1, j, grid,
vis, ans, z,
z_count)
# Left
if (j >= 1 and not vis[i][j - 1] and
grid[i][j - 1] != -1):
grid, vis, ans = dfs(i, j - 1, grid,
vis, ans, z,
z_count)
# Right
if (j < m - 1 and not vis[i][j + 1] and
grid[i][j + 1] != -1):
grid, vis, ans = dfs(i, j + 1, grid,
vis, ans, z,
z_count)
# Unmark the block (unvisited)
vis[i][j] = 0
return grid, vis, ans
# Function to return the count
# of the unique paths
def uniquePaths(grid):
# Total 0s present
z_count = 0
n = len(grid)
m = len(grid[0])
ans = 0
vis = [[0 for j in range(m)]
for i in range(n)]
x = 0
y = 0
for i in range(n):
for j in range(m):
# Count non-obstacle blocks
if grid[i][j] == 0:
z_count += 1
elif (grid[i][j] == 1):
# Starting position
x = i
y = j
grid, vis, ans = dfs(x, y, grid,
vis, ans, 0,
z_count)
return ans
# Driver code
if __name__=='__main__':
grid = [ [ 1, 0, 0, 0 ],
[ 0, 0, 0, 0 ],
[ 0, 0, 2, -1 ] ]
print(uniquePaths(grid))
# This code is contributed by rutvik_56输出：
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