假设给定一个维度为 hx w 的矩阵。该矩阵包含英文字母。我们必须创建另一个包含回文行和列的矩阵，即每一行和每一列都是回文。为此，可以从给定的矩阵中完成任何行和列的排列；但是不能更改任何元素，即不能将“a”更改为“b”。如果可以从给定的矩阵生成回文矩阵，则返回 true；否则，我们返回 false。
因此，如果输入类似于 h = 4, w = 4, mat = {"xxyy", "xyxx", "yxxy", "xyyy"}，那么输出将为真。
脚步
为了解决这个问题，我们将遵循以下步骤 -
Define one map mp Define an array count of size 4. for initialize i := 0, when i < h, update (increase i by 1), do: for initialize j := 0, when j < w, update (increase j by 1), do: (increase tp[mat[i, j]] by 1) for each value val in tp, do: increase count[second value of val mod 4] by 1 check := true if h mod 2 is same as 0 and w mod 2 is same as 0, then: if count[1] + count[2] + count[3] > 0, then: check := false otherwise when h mod 2 is same as 1 and w mod 2 is same as 1, then: if count[1] + count[3] > 1, then: check := false otherwise when count[2] > h / 2 + w / 2, then: check := false Otherwise if count[1] + count[3] > 0, then: check := false otherwise when h mod 2 is same as 1 and count[2] > w / 2, then: check := false otherwise when w mod 2 is same as 1 and count[2] > h / 2, then: check := false return check
示例
让我们看看以下实现以更好地理解 -
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
bool solve(int h, int w, vector<string> mat){
map<char, int> tp;
vector<int> count(4);
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j)
tp[mat[i][j]]++;
}
for (auto val : tp)
count[val.second % 4]++;
bool check = true;
if (h % 2 == 0 && w % 2 == 0) {
if (count[1] + count[2] + count[3] > 0)
check = false;
}
else if (h % 2 == 1 && w % 2 == 1) {
if (count[1]+count[3] > 1)
check = false;
else if (count[2] > h / 2 + w / 2)
check = false;
} else {
if (count[1] + count[3] > 0)
check = false;
else if (h % 2 == 1 && count[2] > w / 2)
check = false;
else if (w % 2 == 1 && count[2] > h / 2)
check = false;
}
return check;
}
int main() {
int h = 4, w = 4;
vector<string> mat = {"xxyy", "xyxx", "yxxy", "xyyy"};
cout<< solve(h, w, mat);
return 0;
}输入
4, 4, {"xxyy", "xyxx", "yxxy", "xyyy"}输出结果1
