给定尺寸为N×M的矩阵m [] []和整数K ，计算XOR(i，j)等于从索引(1，1)到(i，j)的子矩阵所有元素的按位Xor。 ) ，用于矩阵的每个索引。任务是找到具有第K个最大XOR(i，j)值的子矩阵{(1，1)，…，(i，j)} 。如果存在多个这样的子矩阵，则打印最小的子矩阵。
注意：考虑矩阵从(1，1)开始的索引。
例子：
Input: m[][] = {{1, 2}, {2, 3}}, K = 2
Output: 1 2
Explanation:
XOR(1, 1) : m[1][1] = 1
XOR(1, 2): m[1][1] xor m[1][2] = 3
XOR(2, 1): m[1][1] xor m[2][1] = 3
XOR(2, 2): m[1][1] xor m[1][2] xor m[2][1] xor m[2][2] = 2
Hence, the 2nd maximum value is 3 at position [1, 2].
Input: m[][] = {{1, 2, 3}, {2, 2, 1}, {2, 4, 2} }, k = 1
Output: 3 2
方法：想法是使用动态编程找到XOR(i，j)。
- 按xor [i] [j] = xor [i-1] [j] ^ xor [i] [j-1] ^ xor [i-1] [j-1] ^计算按位XOR(i，j) m [i] [j]。
- 将针对各个索引(i，j)获得的XOR(i，j)值存储在Map中。
- 使用大小为K的最小堆找到所有XOR(i，j)值的第K个最大值。
- 使用Map找到XOR(i，j)等于在上述步骤中获得的第K个最大值的最小索引(i，j) 。
下面是上述方法的实现：
C++
// C++ program for above approach
#include
using namespace std;
// Function to print smallest index of
// Kth maximum xors value of submatrices
void smallestPosition(int m[][3], int k, int row)
{
// Dimensions of matrix
int n = row;
int mm = row;
// Stores xors values for every index
int xors[n][mm];
// Min heap to find the
// kth maximum xors value
priority_queue,
greater> minHeap;
// Stores indices for
// corresponding xors vlaues
map> map;
// Traversing matrix to
// calculate xors values
for(int i = 0; i < n; i++)
{
for(int j = 0; j < mm; j++)
{
int a = i - 1 >= 0 ? xors[i - 1][j] : 0;
int b = j - 1 >= 0 ? xors[i][j - 1] : 0;
int c = (i - 1 >= 0 && j - 1 >= 0) ?
xors[i - 1][j - 1] : 0;
xors[i][j] = m[i][j] ^ a ^ b ^ c;
// Insert calculated value
// in Min Heap
minHeap.push(xors[i][j]);
// If size exceeds k
if (minHeap.size() > k)
{
// Remove the minimum
minHeap.pop();
}
// Store smallest index
// containing xors[i][j]
if (map.find(xors[i][j]) == map.end())
map[xors[i][j]] = {i, j};
}
}
// Stores the kth maximum element
int kth_max_e = minHeap.top();
// Print the required index
cout << (map[kth_max_e][0] + 1) << " "
<< (map[kth_max_e][1] + 1);
}
// Driver Code
int main()
{
int m[][3] = { { 1, 2, 3 },
{ 2, 2, 1 },
{ 2, 4, 2 } };
int k = 1;
// Function call
smallestPosition(m, k, 3);
}
// This code is contributed by grand_master Java
// Java Program for above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to print smallest index of
// Kth maximum Xor value of submatrices
static void smallestPosition(int m[][], int k)
{
// Dimensions of matrix
int n = m.length;
int mm = m[0].length;
// Stores XOR values for every index
int[][] xor = new int[n][mm];
// Min heap to find the
// kth maximum XOR value
PriorityQueue minHeap
= new PriorityQueue<>();
// Stores indices for
// corresponding XOR vlaues
Map map
= new HashMap<>();
// Traversing matrix to
// calculate XOR values
for (int i = 0; i < n; i++) {
for (int j = 0; j < mm; j++) {
int a = i - 1 >= 0
? xor[i - 1][j]
: 0;
int b = j - 1 >= 0
? xor[i][j - 1]
: 0;
int c = (i - 1 >= 0 && j - 1 >= 0)
? xor[i - 1][j - 1]
: 0;
xor[i][j] = m[i][j] ^ a ^ b ^ c;
// Insert calculated value
// in Min Heap
minHeap.add(xor[i][j]);
// If size exceeds k
if (minHeap.size() > k) {
// Remove the minimum
minHeap.poll();
}
// Store smallest index
// containing xor[i][j]
if (!map.containsKey(xor[i][j]))
map.put(xor[i][j],
new int[] { i, j });
}
}
// Stores the kth maximum element
int kth_max_e = minHeap.poll();
// Print the required index
System.out.println(
(map.get(kth_max_e)[0] + 1)
+ " " + (map.get(kth_max_e)[1] + 1));
}
// Driver Code
public static void main(String[] args)
{
int m[][] = { { 1, 2, 3 },
{ 2, 2, 1 },
{ 2, 4, 2 } };
int k = 1;
// Function call
smallestPosition(m, k);
}
} Python3
# Python3 Program for the
# above approach
# Function to print smallest index of
# Kth maximum Xor value of submatrices
def smallestPosition(m, k) :
# Dimensions of matrix
n = len(m)
mm = len(m[1])
# Stores XOR values for
# every index
xor = [[0 for i in range(mm)] for j in range(n)]
# Min heap to find the
# kth maximum XOR value
minHeap = []
# Stores indices for
# corresponding XOR vlaues
Map = {}
# Traversing matrix to
# calculate XOR values
for i in range(n) :
for j in range(mm) :
if i - 1 >= 0 :
a = xor[i - 1][j]
else :
a = 0
if j - 1 >= 0 :
b = xor[i][j - 1]
else :
b = 0
if i - 1 >= 0 and j - 1 >= 0 :
c = xor[i - 1][j - 1]
else :
c = 0
xor[i][j] = m[i][j] ^ a ^ b ^ c
# Insert calculated value
# in Min Heap
minHeap.append(xor[i][j])
minHeap.sort()
# If size exceeds k
if (len(minHeap) > k) :
# Remove the minimum
del minHeap[0]
# Store smallest index
# containing xor[i,j]
if xor[i][j] not in Map :
Map[xor[i][j]] = [i, j]
minHeap.sort()
# Stores the kth maximum
# element
kth_max_e = minHeap[0]
# Print the required index
print((Map[kth_max_e][0] + 1), (Map[kth_max_e][1] + 1))
# Driver code
m = [[1, 2, 3],
[2, 2, 1],
[2, 4, 2]]
k = 1
# Function call
smallestPosition(m, k)
# This code is contributed by divyesh072019C#
// C# Program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print smallest index of
// Kth maximum Xor value of submatrices
static void smallestPosition(int [,]m,
int k)
{
// Dimensions of matrix
int n = m.GetLength(0);
int mm = m.GetLength(1);
// Stores XOR values for
// every index
int[,] xor = new int[n, mm];
// Min heap to find the
// kth maximum XOR value
List minHeap =
new List();
// Stores indices for
// corresponding XOR vlaues
Dictionary map = new Dictionary();
// Traversing matrix to
// calculate XOR values
for (int i = 0; i < n; i++)
{
for (int j = 0; j < mm; j++)
{
int a = i - 1 >= 0 ?
xor[i - 1, j] : 0;
int b = j - 1 >= 0 ?
xor[i, j - 1] : 0;
int c = (i - 1 >= 0 &&
j - 1 >= 0) ?
xor[i - 1, j - 1] : 0;
xor[i, j] = m[i, j] ^
a ^ b ^ c;
// Insert calculated value
// in Min Heap
minHeap.Add(xor[i, j]);
minHeap.Sort();
// If size exceeds k
if (minHeap.Count > k)
{
// Remove the minimum
minHeap.RemoveAt(0);
}
// Store smallest index
// containing xor[i,j]
if (!map.ContainsKey(xor[i, j]))
map.Add(xor[i, j],
new int[] {i, j});
}
}
minHeap.Sort();
// Stores the kth maximum
// element
int kth_max_e = minHeap[0];
// Print the required index
Console.WriteLine((map[kth_max_e][0] + 1) +
" " + (map[kth_max_e][1] + 1));
}
// Driver Code
public static void Main(String[] args)
{
int [,]m = {{1, 2, 3},
{2, 2, 1},
{2, 4, 2}};
int k = 1;
// Function call
smallestPosition(m, k);
}
}
// This code is contributed by Amit Katiyar 输出：
3 2时间复杂度： O(N * M * log K)
辅助空间： O(N * M)