包含相交同心子矩阵的矩阵中的最大值
假设一个大小为NXN的矩阵包含以(x i , y i )为中心的同心方阵子矩阵，其中 x i是第 i 个同心方阵中心的行数，y i是第 i 个同心方阵中心的列数。同心方阵的形式为：
0 0 0 0 0 0 0 0 0
0 1 1 1 1 1 1 1 0
0 1 . . . . . 1 0
0 1 . b b b . 1 0
0 1 . b a b . 1 0
0 1 . b b b . 1 0
0 1 . . . . . 1 0
0 1 1 1 1 1 1 1 0
0 0 0 0 0 0 0 0 0其中 a 为中心，b 为 a – 1，随着行或列的增加，该值会减小。
由于存在多个这样的子矩阵，因此存在属于多个这样的子矩阵的一部分的单元。这些单元格的值将等于相交子矩阵的值之和。给定N, m, (x i , y i , a i ) 的值，其中 1 <= i <= m 和 a i是第 i 个同心子矩阵中心的值。任务是在包含子矩阵的矩阵中找到最大值。
所以，经过
例子：
Input : N = 10, m = 2
(x1, y1, a1) = (3, 3, 3)
(x2, y2, a2) = (7, 7, 4)
Output : 4
Maxtrix that will be form:
Input : N = 10, m = 1
(x1, y1, a1) = (4, 5, 6)
Output : 6这个想法是制作一个二维矩阵mat[][]并找到每个单元格的值，包括具有多个同心子矩阵的交集的单元格。现在，观察每个单元格的值可以通过 max(0, a – max(p – x i , q – y i )) 找到，其中 a 是第 i 个同心子矩阵中心的值，p 是行单元格的编号，q 是单元格的列数，(x i , y i ) 是第 i 个同心子矩阵中心的中心位置。所以，在找到矩阵 mat[][] 之后，我们将遍历矩阵以找到矩阵中的最大值。
下面是这种方法的 C++ 实现：
C++
//C++ Program to find the maximum value in a matrix
//which contain intersecting concentric submatrix
#include
using namespace std;
#define MAXN 100
// Return the maximum value in intersecting
// concentric submatrix.
int maxValue(int n, int m, int x[], int y[], int a[])
{
int c[MAXN][MAXN] = { 0 };
// For each center of concentric sub-matrix.
for (int i = 0; i < m; ++i) {
// for each row
for (int p = 0; p < n; ++p) {
// for each column
for (int q = 0; q < n; ++q) {
// finding x distance.
int dx = abs(p - x[i]);
// finding y distance.
int dy = abs(q - y[i]);
// maximum of x distance and y distance
int d = max(dx, dy);
// assigning the value.
c[p][q] += max(0, a[i] - d);
}
}
}
// Finding the maximum value in the formed matrix.
int res = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
res = max(res, c[i][j]);
}
}
return res;
}
// Driven Program
int main()
{
int n = 10;
int m = 2;
int x[] = { 3, 7 };
int y[] = { 3, 7 };
int a[] = { 4, 3 };
cout << maxValue(n, m, x, y, a) << endl;
return 0;
} Java
// Java Program to find the
// maximum value in a matrix
// which contain intersecting
// concentric submatrix
import java.io.*;
class GFG
{
static int MAXN = 100;
// Return the maximum value
// in intersecting
// concentric submatrix.
static int maxValue(int n, int m,
int x[], int y[],
int a[])
{
int c[][] = new int[MAXN][MAXN];
// For each center of
// concentric sub-matrix.
for (int i = 0; i < m; ++i)
{
// for each row
for (int p = 0; p < n; ++p)
{
// for each column
for (int q = 0; q < n; ++q)
{
// finding x distance.
int dx = Math.abs(p - x[i]);
// finding y distance.
int dy = Math.abs(q - y[i]);
// maximum of x distance
// and y distance
int d = Math.max(dx, dy);
// assigning the value.
c[p][q] += Math.max(0, a[i] - d);
}
}
}
// Finding the maximum
// value in the formed matrix.
int res = 0;
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
{
res = Math.max(res, c[i][j]);
}
}
return res;
}
// Driven Code
public static void main (String[] args)
{
int n = 10;
int m = 2;
int x[] = { 3, 7 };
int y[] = { 3, 7 };
int a[] = { 4, 3 };
System.out.println(maxValue(n, m, x,
y, a));
}
}
// This code is contributed by anuj_67.Python 3
# Python 3 Program to find the maximum
# value in a matrix which contain
# intersecting concentric submatrix
MAXN = 100
# Return the maximum value in intersecting
# concentric submatrix.
def maxValue( n, m, x, y, a):
c = [[0 for x in range(MAXN)]
for y in range(MAXN)]
# For each center of concentric sub-matrix.
for i in range( m):
# for each row
for p in range(n) :
# for each column
for q in range( n) :
# finding x distance.
dx = abs(p - x[i])
# finding y distance.
dy = abs(q - y[i])
# maximum of x distance and y distance
d = max(dx, dy)
# assigning the value.
c[p][q] += max(0, a[i] - d)
# Finding the maximum value in
# the formed matrix.
res = 0
for i in range(n) :
for j in range(n) :
res = max(res, c[i][j])
return res
# Driver Code
if __name__ == "__main__":
n = 10
m = 2
x = [ 3, 7 ]
y = [ 3, 7 ]
a = [ 4, 3 ]
print(maxValue(n, m, x, y, a))
# This code is contributed by ita_cC#
// C# Program to find the maximum
// value in a matrix which contain
// intersecting concentric submatrix
using System;
class GFG
{
static int MAXN = 100;
// Return the maximum value in intersecting
// concentric submatrix.
static int maxValue(int n, int m,
int[] x, int[] y,
int[] a)
{
int[,] c = new int[MAXN, MAXN];
// For each center of
// concentric sub-matrix.
for (int i = 0; i < m; ++i)
{
// for each row
for (int p = 0; p < n; ++p)
{
// for each column
for (int q = 0; q < n; ++q)
{
// finding x distance.
int dx = Math.Abs(p - x[i]);
// finding y distance.
int dy = Math.Abs(q - y[i]);
// maximum of x distance
// and y distance
int d = Math.Max(dx, dy);
// assigning the value.
c[p,q] += Math.Max(0, a[i] - d);
}
}
}
// Finding the maximum
// value in the formed matrix.
int res = 0;
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
{
res = Math.Max(res, c[i, j]);
}
}
return res;
}
// Driver Code
public static void Main ()
{
int n = 10;
int m = 2;
int[] x = { 3, 7 };
int[] y = { 3, 7 };
int[] a = { 4, 3 };
Console.Write(maxValue(n, m, x, y, a));
}
}Javascript
输出：
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