我们得到一个矩阵，其中每个单元格中包含不同的值。我们的目标是找到矩阵中位置的最小集合，以便可以从集合中的位置开始遍历整个矩阵。
我们可以在以下条件下遍历矩阵：
- 我们只能移动到那些包含小于或等于当前单元格值的邻居。小区的邻居被定义为与给定小区共享一侧的小区。
例子：
Input : 1 2 3
2 3 1
1 1 1
Output : 1 1
0 2
If we start from 1, 1 we can cover 6
vertices in the order (1, 1) -> (1, 0) -> (2, 0)
-> (2, 1) -> (2, 2) -> (1, 2). We cannot cover
the entire matrix with this vertex. Remaining
vertices can be covered (0, 2) -> (0, 1) -> (0, 0).
Input : 3 3
1 1
Output : 0 1
If we start from 0, 1, we can traverse
the entire matrix from this single vertex
in this order (0, 0) -> (0, 1) -> (1, 1) -> (1, 0).
Traversing the matrix in this order
satisfies all the conditions stated above.从上面的例子中，我们可以很容易地确定，为了使用最少数量的仓位，我们必须从具有最高单元格值的仓位开始。因此，我们选择矩阵中包含最高值的位置。我们在单独的数组中取具有最高值的顶点。我们在每个顶点执行 DFS，从
最高值。如果我们在 dfs 期间遇到任何未访问的顶点，那么我们必须将该顶点包含在我们的集合中。处理完所有单元格后，该集合将包含所需的顶点。
这是如何运作的？
我们需要访问所有顶点并达到我们必须从它们开始的最大值。如果两个最大值不相邻，则必须同时选取它们。如果两个最大值相邻，则可以选择其中任何一个，因为允许移动到相等值的邻居。
C++
// C++ program to find minimum initial
// vertices to reach whole matrix.
#include
using namespace std;
const int MAX = 100;
// (n, m) is current source cell from which
// we need to do DFS. N and M are total no.
// of rows and columns.
void dfs(int n, int m, bool visit[][MAX],
int adj[][MAX], int N, int M)
{
// Marking the vertex as visited
visit[n][m] = 1;
// If below neighbor is valid and has
// value less than or equal to current
// cell's value
if (n + 1 < N &&
adj[n][m] >= adj[n + 1][m] &&
!visit[n + 1][m])
dfs(n + 1, m, visit, adj, N, M);
// If right neighbor is valid and has
// value less than or equal to current
// cell's value
if (m + 1 < M &&
adj[n][m] >= adj[n][m + 1] &&
!visit[n][m + 1])
dfs(n, m + 1, visit, adj, N, M);
// If above neighbor is valid and has
// value less than or equal to current
// cell's value
if (n - 1 >= 0 &&
adj[n][m] >= adj[n - 1][m] &&
!visit[n - 1][m])
dfs(n - 1, m, visit, adj, N, M);
// If left neighbor is valid and has
// value less than or equal to current
// cell's value
if (m - 1 >= 0 &&
adj[n][m] >= adj[n][m - 1] &&
!visit[n][m - 1])
dfs(n, m - 1, visit, adj, N, M);
}
void printMinSources(int adj[][MAX], int N, int M)
{
// Storing the cell value and cell indices
// in a vector.
vector > > x;
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
x.push_back(make_pair(adj[i][j],
make_pair(i, j)));
// Sorting the newly created array according
// to cell values
sort(x.begin(), x.end());
// Create a visited array for DFS and
// initialize it as false.
bool visit[N][MAX];
memset(visit, false, sizeof(visit));
// Applying dfs for each vertex with
// highest value
for (int i = x.size()-1; i >=0 ; i--)
{
// If the given vertex is not visited
// then include it in the set
if (!visit[x[i].second.first][x[i].second.second])
{
cout << x[i].second.first << " "
<< x[i].second.second << endl;
dfs(x[i].second.first, x[i].second.second,
visit, adj, N, M);
}
}
}
// Driver code
int main()
{
int N = 2, M = 2;
int adj[N][MAX] = {{3, 3},
{1, 1}};
printMinSources(adj, N, M);
return 0;
} Python3
# Python3 program to find minimum initial
# vertices to reach whole matrix
MAX = 100
# (n, m) is current source cell from which
# we need to do DFS. N and M are total no.
# of rows and columns.
def dfs(n, m, visit, adj, N, M):
# Marking the vertex as visited
visit[n][m] = 1
# If below neighbor is valid and has
# value less than or equal to current
# cell's value
if (n + 1 < N and
adj[n][m] >= adj[n + 1][m] and
not visit[n + 1][m]):
dfs(n + 1, m, visit, adj, N, M)
# If right neighbor is valid and has
# value less than or equal to current
# cell's value
if (m + 1 < M and
adj[n][m] >= adj[n][m + 1] and
not visit[n][m + 1]):
dfs(n, m + 1, visit, adj, N, M)
# If above neighbor is valid and has
# value less than or equal to current
# cell's value
if (n - 1 >= 0 and
adj[n][m] >= adj[n - 1][m] and
not visit[n - 1][m]):
dfs(n - 1, m, visit, adj, N, M)
# If left neighbor is valid and has
# value less than or equal to current
# cell's value
if (m - 1 >= 0 and
adj[n][m] >= adj[n][m - 1] and
not visit[n][m - 1]):
dfs(n, m - 1, visit, adj, N, M)
def printMinSources(adj, N, M):
# Storing the cell value and cell
# indices in a vector.
x = []
for i in range(N):
for j in range(M):
x.append([adj[i][j], [i, j]])
# Sorting the newly created array according
# to cell values
x.sort()
# Create a visited array for DFS and
# initialize it as false.
visit = [[False for i in range(MAX)]
for i in range(N)]
# Applying dfs for each vertex with
# highest value
for i in range(len(x) - 1, -1, -1):
# If the given vertex is not visited
# then include it in the set
if (not visit[x[i][1][0]][x[i][1][1]]):
print('{} {}'.format(x[i][1][0],
x[i][1][1]))
dfs(x[i][1][0],
x[i][1][1],
visit, adj, N, M)
# Driver code
if __name__=='__main__':
N = 2
M = 2
adj = [ [ 3, 3 ], [ 1, 1 ] ]
printMinSources(adj, N, M)
# This code is contributed by rutvik_56Javascript
输出：
0 1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。