C++程序以找出将所有单元格转换为黑色所需的迭代次数

假设，我们有一个包含两种类型单元格的网格；黑细胞和白细胞。黑色单元格表示为“#”，白色单元格表示为“.”。网格以字符串数组形式提供给我们。现在，我们必须执行以下操作。

• 我们将每个白色单元格转换为黑色单元格，并与黑色单元格共享一侧。我们执行此操作，直到网格的每个单元格都变成黑色。

• 我们计算将网格的所有单元格转换为黑色所需的迭代次数。一开始的网格必须包含一个黑色单元格。

因此，如果输入类似于 h = 4, w = 4, grid = {"#..." , ".#.." , "....", "...#"}

那么输出将为3。

需要3次迭代才能转换所有单元格为黑色。

步骤

为了解决这个问题，我们将按照以下步骤操作 -

Define an array dx of size: 4 containing := { 1, 0, - 1, 0 }
Define an array dy of size: 4 containing := { 0, 1, 0, - 1 }
Define one 2D array distance
Define one queue q that contain integer pairs
for initialize i := 0, when i < h, update (increase i by 1), do:
for initialize j := 0, when j < w, update (increase j by 1), do:
if grid[i, j] is same as '#', then:
distance[i, j] := 0
insert one pair(i, j) into q
while (not q is empty), do:
first element of auto now = q
delete element from q
for initialize dir := 0, when dir < 4, update (increase dir by 1), do:
cx := first value of now + dx[dir]
cy := second value of now + dy[dir]
if cx = h or cy = w, then:
if distance[cx, cy] is same as -1, then:
distance[cx, cy] := distance[first value of now, second value of now] + 1
insert one pair (cx, cy) into q
ans := 0
for initialize i := 0, when i < h, update (increase i by 1), do:
for initialize j := 0, when j < w, update (increase j by 1), do:
ans := maximum of ans and distance[i, j]
print(ans)

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示例

让我们看下面的实现以获得更好的理解 −

#include
using namespace std;

void solve(int h, int w, vector grid){
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };
vector distance(h, vector(w, -1));
queue q;
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (grid[i][j] == '#') {
distance[i][j] = 0;
q.push(pair(i,j));
}
}
}
while (!q.empty()) {
auto now = q.front();
q.pop();
for (int dir = 0; dir < 4; dir++) {
int cx = now.first + dx[dir];
int cy = now.second + dy[dir];
if (cx = h || cy = w) continue;
if (distance[cx][cy] == -1) {
distance[cx][cy] = distance[now.first][now.second] + 1;
q.push(pair (cx, cy));
}
}
}
int ans = 0; for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
ans = max(ans, distance[i][j]);
}
}
cout