假設我們有一個 n x n 矩陣。矩陣中的每個元素都是唯一的，並且是 1 到 n2 之間的整數。現在我們可以以任意數量和任意順序執行以下操作。
我們選擇矩陣中的任兩個整數 x 和 y，其中 (1 ≤ x
我們選擇矩陣中的任兩個整數 x 和 y，其中 (1 ≤ x
我們必須注意 x y ≤ k 且這些值不能出現在相同的行和列中。
我們必須找出透過執行操作可以獲得的唯一矩陣的數量。
因此，如果輸入類似於n = 3, k = 15, mat = {{4, 3, 6}, {5, 9, 7}, {1, 2, 8}}，那麼輸出將為36。
例如，選取的兩個值是 x = 3 和 y = 5。如果交換列，所得矩陣將是 -
3 4 6 9 5 7 2 1 8
透過這種方式可以獲得36個這樣的獨特矩陣。
為了解決這個問題，我們將遵循以下步驟-
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)讓我們看看以下實現，以便更好地理解-
#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;
}3, 15, {{4, 3, 6}, {5, 9, 7}, {1, 2, 8}}36
以上是C++程式以找出透過交換行和列可以產生的唯一矩陣的數量的詳細內容。更多資訊請關注PHP中文網其他相關文章！
