计算二进制矩阵中 1 和 0 的集合数量
原文： https://www.geeksforgeeks.org/counting-sets-of-1s-and-0s-in-a-binary-matrix/
给定一个n×m的二进制矩阵，计算可以在行或列中形成一个或多个相同值的集合的数量。
示例：
Input: 1 0 1
0 1 0
Output: 8
Explanation: There are six one-element sets
(three 1s and three 0s). There are two two-
element sets, the first one consists of the
first and the third cells of the first row.
The second one consists of the first and the
third cells of the second row.
Input: 1 0
1 1
Output: 6
x个元素的非空子集数量为2 ^ x –1。我们遍历每一行并计算 1 和 0 的单元格数。 对于u个零和v个一，总集合为2 ^ u – 1 + 2 ^ v –1。然后遍历所有列并计算相同的值并计算总和。 我们最终从总和中减去m x n，因为单个元素被考虑了两次。
CPP
// CPP program to compute number of sets
// in a binary matrix.
#include <bits/stdc++.h>
using namespace std;
const int m = 3; // no of columns
const int n = 2; // no of rows
// function to calculate the number of
// non empty sets of cell
long long countSets(int a[n][m])
{
// stores the final answer
long long res = 0;
// traverses row-wise
for (int i = 0; i < n; i++)
{
int u = 0, v = 0;
for (int j = 0; j < m; j++)
a[i][j] ? u++ : v++;
res += pow(2,u)-1 + pow(2,v)-1;
}
// traverses column wise
for (int i = 0; i < m; i++)
{
int u = 0, v = 0;
for (int j = 0; j < n; j++)
a[j][i] ? u++ : v++;
res += pow(2,u)-1 + pow(2,v)-1;
}
// at the end subtract n*m as no of
// single sets have been added twice.
return res-(n*m);
}
// driver program to test the above function.
int main() {
int a[][3] = {(1, 0, 1),
(0, 1, 0)};
cout << countSets(a);
return 0;
}
Java
// Java program to compute number of sets
// in a binary matrix.
class GFG {
static final int m = 3; // no of columns
static final int n = 2; // no of rows
// function to calculate the number of
// non empty sets of cell
static long countSets(int a[][]) {
// stores the final answer
long res = 0;
// traverses row-wise
for (int i = 0; i < n; i++) {
int u = 0, v = 0;
for (int j = 0; j < m; j++) {
if (a[i][j] == 1)
u++;
else
v++;
}
res += Math.pow(2, u) - 1 + Math.pow(2, v) - 1;
}
// traverses column wise
for (int i = 0; i < m; i++) {
int u = 0, v = 0;
for (int j = 0; j < n; j++) {
if (a[j][i] == 1)
u++;
else
v++;
}
res += Math.pow(2, u) - 1 + Math.pow(2, v) - 1;
}
// at the end subtract n*m as no of
// single sets have been added twice.
return res - (n * m);
}
// Driver code
public static void main(String[] args) {
int a[][] = {{1, 0, 1}, {0, 1, 0}};
System.out.print(countSets(a));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to compute number of sets
# in a binary matrix.
m = 3 # no of columns
n = 2 # no of rows
# function to calculate the number of
# non empty sets of cell
def countSets(a):
# stores the final answer
res = 0
# traverses row-wise
for i in range(n):
u = 0
v = 0
for j in range(m):
if a[i][j]:
u += 1
else:
v += 1
res += pow(2, u) - 1 + pow(2, v) - 1
# traverses column wise
for i in range(m):
u = 0
v = 0
for j in range(n):
if a[j][i]:
u += 1
else:
v += 1
res += pow(2, u) - 1 + pow(2, v) - 1
# at the end subtract n*m as no of
# single sets have been added twice.
return res - (n*m)
# Driver program to test the above function.
a = [[1, 0, 1],[0, 1, 0]]
print(countSets(a))
# This code is conributed by shubhamsingh10
C#
// C# program to compute number of
// sets in a binary matrix.
using System;
class GFG {
static int m = 3; // no of columns
static int n = 2; // no of rows
// function to calculate the number of
// non empty sets of cell
static long countSets(int [,]a)
{
// stores the final answer
long res = 0;
// Traverses row-wise
for (int i = 0; i < n; i++)
{
int u = 0, v = 0;
for (int j = 0; j < m; j++)
{
if (a[i,j] == 1)
u++;
else
v++;
}
res += (long)(Math.Pow(2, u) - 1
+ Math.Pow(2, v)) - 1;
}
// Traverses column wise
for (int i = 0; i < m; i++)
{
int u = 0, v = 0;
for (int j = 0; j < n; j++)
{
if (a[j,i] == 1)
u++;
else
v++;
}
res += (long)(Math.Pow(2, u) - 1
+ Math.Pow(2, v)) - 1;
}
// at the end subtract n*m as no of
// single sets have been added twice.
return res - (n * m);
}
// Driver code
public static void Main()
{
int [,]a = {{1, 0, 1}, {0, 1, 0}};
Console.WriteLine(countSets(a));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to compute
// number of sets
// in a binary matrix.
// no of columns
$m = 3;
// no of rows
$n = 2;
// function to calculate the number
// of non empty sets of cell
function countSets($a)
{
global $m, $n;
// stores the final answer
$res = 0;
// traverses row-wise
for ($i = 0; $i < $n; $i++)
{
$u = 0; $v = 0;
for ( $j = 0; $j < $m; $j++)
$a[$i][$j] ? $u++ : $v++;
$res += pow(2, $u) - 1 + pow(2, $v) - 1;
}
// traverses column wise
for ($i = 0; $i < $m; $i++)
{
$u = 0;$v = 0;
for ($j = 0; $j < $n; $j++)
$a[$j][$i] ? $u++ : $v++;
$res += pow(2, $u) - 1 +
pow(2, $v) - 1;
}
// at the end subtract
// n*m as no of single
// sets have been added
// twice.
return $res-($n*$m);
}
// Driver Code
$a = array(array(1, 0, 1),
array(0, 1, 0));
echo countSets($a);
// This code is contributed by anuj_67\.
?>
Output:
8
时间复杂度：O(n * m)。
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