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

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

WBOY
Release: 2023-09-01 11:53:04
forward
1005 people have browsed it

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
Copy after login

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)
Copy after login

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;
}
Copy after login

Input

3, 15, {{4, 3, 6}, {5, 9, 7}, {1, 2, 8}}
Copy after login

Output

36
Copy after login

The above is the detailed content of C++ program to find the number of unique matrices that can be generated by swapping rows and columns. For more information, please follow other related articles on the PHP Chinese website!

Related labels:
source:tutorialspoint.com
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template