给定一个最初为空的N * N板并执行一系列查询，每个查询都由两个整数X和Y组成，其中绘制了单元格(X，Y)。任务是打印查询的编号，此后在板上将显示一个3 * 3的正方形，并绘制了所有单元格。
如果在处理完所有查询后没有这样的正方形，则打印-1 。
例子：
Input: N = 4, q = {{1, 1}, {1, 2}, {1, 3}, {2, 2}, {2, 3}, {1, 4}, {2, 4}, {3, 4}, {3, 2}, {3, 3}, {4, 1}}
Output: 10
After the 10th move, there exists a 3X3 square, from (1, 1) to (3, 3) (1-based indexing).
Input: N = 3, q = {(1, 1), {1, 2}, {1, 3}}
Output: -1
方法：这里的一个重要观察结果是，每次我们为一个正方形着色时，它都可以以9种不同方式(3 * 3正方形的任何单元)成为所需正方形的一部分。对于每种可能性，请检查当前单元格是否是所有9个单元格都已绘制的正方形的一部分。如果满足条件，则打印到目前为止已处理的查询数，否则在处理完所有查询后打印-1。
下面是上述方法的实现：
CPP
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true
// if the coordinate is inside the grid
bool valid(int x, int y, int n)
{
if (x < 1 || y < 1 || x > n || y > n)
return false;
return true;
}
// Function that returns the count of squares
// that are coloured in the 3X3 square
// with (cx, cy) being the top left corner
int count(int n, int cx, int cy,
vector >& board)
{
int ct = 0;
// Iterate through 3 rows
for (int x = cx; x <= cx + 2; x++)
// Iterate through 3 columns
for (int y = cy; y <= cy + 2; y++)
// Check if the current square
// is valid and coloured
if (valid(x, y, n))
ct += board[x][y];
return ct;
}
// Function that returns the query
// number after which the grid will
// have a 3X3 coloured square
int minimumMoveSquare(int n, int m,
vector > moves)
{
int x, y;
vector > board(n + 1);
// Initialize all squares to be uncoloured
for (int i = 1; i <= n; i++)
board[i].resize(n + 1, false);
for (int i = 0; i < moves.size(); i++) {
x = moves[i].first;
y = moves[i].second;
// Mark the current square as coloured
board[x][y] = true;
// Check if any of the 3X3 squares
// which contains the current square
// is fully coloured
for (int cx = x - 2; cx <= x; cx++)
for (int cy = y - 2; cy <= y; cy++)
if (count(n, cx, cy, board) == 9)
return i + 1;
}
return -1;
}
// Driver code
int main()
{
int n = 3;
vector > moves = { { 1, 1 },
{ 1, 2 },
{ 1, 3 } };
int m = moves.size();
cout << minimumMoveSquare(n, m, moves);
return 0;
} Python3
# Python3 implementation of the approach
# Function that returns True
# if the coordinate is inside the grid
def valid(x, y, n):
if (x < 1 or y < 1 or x > n or y > n):
return False;
return True;
# Function that returns the count of squares
# that are coloured in the 3X3 square
# with (cx, cy) being the top left corner
def count(n, cx, cy, board):
ct = 0;
# Iterate through 3 rows
for x in range(cx, cx + 3):
# Iterate through 3 columns
for y in range(cy + 3):
# Check if the current square
# is valid and coloured
if (valid(x, y, n)):
ct += board[x][y];
return ct;
# Function that returns the query
# number after which the grid will
# have a 3X3 coloured square
def minimumMoveSquare(n, m, moves):
x = 0
y = 0
board=[[] for i in range(n + 1)]
# Initialize all squares to be uncoloured
for i in range(1, n + 1):
board[i] = [False for i in range(n + 1)]
for i in range(len(moves)):
x = moves[i][0];
y = moves[i][1];
# Mark the current square as coloured
board[x][y] = True;
# Check if any of the 3X3 squares
# which contains the current square
# is fully coloured
for cx in range(x - 2, x + 1 ):
for cy in range(y - 2, y + 1):
if (count(n, cx, cy, board) == 9):
return i + 1;
return -1;
# Driver code
if __name__=='__main__':
n = 3;
moves = [[ 1, 1 ],[ 1, 2 ], [ 1, 3 ]]
m = len(moves)
print(minimumMoveSquare(n, m, moves))
# This code is contributed by rutvik_56.输出：
-1