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C++ Program to find out the number of unique matrices that can be generated by swapping rows and columns
Suppose, we have a n x n matrix. Each element in the matrix is unique and is an integer number between 1 and n2. Now we can perform the operations below in any amount and any order.
We pick any two integers x and y that are in the matrix, where (1 ≤ x < y ≤ n) and swap the columns containing x and y.
We pick any two integers x and y that are in the matrix, where (1 ≤ x < y ≤ n) and swap the rows containing x and y.
We have to note that x + y ≤ k and the values must not be present in the same rows and columns.
We have to find out the number of unique matrices that can be obtained by performing the operations.
So, if the input is like n = 3, k = 15, mat = {{4, 3, 6}, {5, 9, 7}, {1, 2, 8}}, then the output will be 36.
For example, the two values picked are x = 3 and y = 5. The resultant matrix if the columns are swapped will be −
3 4 6 9 5 7 2 1 8
36 such unique matrices can be obtained this way.
To solve this, we will follow these steps −
Define a function dfs(), this will take k, arrays ver and visited, one stack s. if visited[k] is non-zero, then: return visited[k] := true insert k into s for initialize iterator j := start of ver[k], when j is not equal to last element of ver[k], update (increase j by 1), do: dfs(*j, ver, visited, s) Define an array f of size: 51. f[0] := 1 for initialize i := 1, when i <= 50, update (increase i by 1), do: f[i] := (i * f[i - 1]) mod modval Define an array e of size n Define an array pk of size n for initialize i := 0, when i < n, update (increase i by 1), do: for initialize j := i + 1, when j < n, update (increase j by 1), do: chk := 0 for initialize l := 0, when l < n, update (increase l by 1), do: if (mat[i, l] + mat[j, l]) > k, then: chk := 1 Come out from the loop if chk is same as 0, then: insert j at the end of pk[i] insert i at the end of pk[j] chk := 0 for initialize l := 0, when l < n, update (increase l by 1), do: if (mat[l, i] + mat[l, j]) > k, then: chk := 1 Come out from the loop if chk is same as 0, then: insert j at the end of e[i] insert i at the end of e[j] resa := 1, resb = 1 Define an array v1 of size: n and v2 of size: n. for initialize i := 0, when i < n, update (increase i by 1), do: v1[i] := false v2[i] := false for initialize i := 0, when i < n, update (increase i by 1), do: Define one stack s. if not v1[i] is non-zero, then: dfs(i, pk, v1, s) if not s is empty, then: resa := resa * (f[size of s]) resa := resa mod modval for initialize i := 0, when i < n, update (increase i by 1), do: Define one stack s if not v2[i] is non-zero, then: dfs(i, e, v2, s) if not s is empty, then: resb := resb * (f[size of s]) resb := resb mod modval print((resa * resb) mod modval)
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
#define modval 998244353
const int INF = 1e9;
void dfs(int k, vector<int> ver[], bool visited[], stack<int> &s) {
if(visited[k])
return;
visited[k] = true;
s.push(k);
for(vector<int> :: iterator j = ver[k].begin(); j!=ver[k].end(); j++)
dfs(*j, ver, visited, s);
}
void solve(int n, int k, vector<vector<int>> mat) {
int f[51];
f[0] = 1;
for(int i = 1; i <= 50; i++) {
f[i] = (i * f[i-1]) % modval;
}
vector<int> e[n];
vector<int> pk[n];
for(int i = 0; i < n; i++) {
for(int j = i + 1;j < n; j++) {
int chk = 0;
for(int l = 0; l < n; l++){
if((mat[i][l] + mat[j][l]) > k) {
chk = 1;
break;
}
}
if(chk==0) {
pk[i].push_back(j);
pk[j].push_back(i);
}
chk = 0;
for(int l = 0;l < n; l++) {
if((mat[l][i] + mat[l][j]) > k){
chk = 1;
break;
}
}
if(chk == 0) {
e[i].push_back(j);
e[j].push_back(i);
}
}
}
int resa = 1, resb = 1;
bool v1[n], v2[n];
for(int i = 0; i < n; i++) {
v1[i] = false;
v2[i] = false;
}
for(int i = 0;i < n; i++) {
stack<int> s;
if(!v1[i]) {
dfs(i, pk, v1, s);
if(!s.empty()) {
resa *= (f[s.size()]) % modval;
resa %= modval;
}
}
}
for(int i = 0 ;i < n; i++) {
stack<int> s;
if(!v2[i]){
dfs(i, e, v2, s);
if(!s.empty()) {
resb *= (f[s.size()]) % modval;
resb %= modval;
}
}
}
cout<< (resa * resb) % modval;
}
int main() {
int n = 3, k = 15;
vector<vector<int>> mat = {{4, 3, 6}, {5, 9, 7}, {1, 2, 8}};
solve(n, k, mat);
return 0;
}Input
3, 15, {{4, 3, 6}, {5, 9, 7}, {1, 2, 8}}Output
36
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