C++ program to find the number of unique matrices that can be generated by swapping rows and columns

Suppose we have an n x n matrix. Each element in the matrix is ​​unique and an integer between 1 and n². Now we can perform the following operations in any number and in any order.

• We choose any two integers x and y in the matrix where (1 ≤ x

• We choose any two integers x and y in the matrix where (1 ≤ x

• We must note that x y ≤ k and these values ​​cannot appear in the same row and column.

We have to find out the number of unique matrices that can be obtained by performing the operation.

So if the input is something 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 ​​selected are x = 3 and y = 5. If you swap the columns, the resulting matrix will be -

3 4 6
9 5 7
2 1 8

In this way you can get 36 such unique matrices.

To solve this problem we will follow the following 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 for 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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