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法律研究检查两个图像在旋转和转换后是否匹配的 C++ 程序
假设有两个 n * n 像素的正方形图像 first 和 second。像素可以是黑色或白色。这些图像以矩阵表示的形式给出，如果一个像素是黑色，则表示为“x”，如果它是白色，则表示为“.”。我们必须检查第二个图像在旋转 90° 和平移后是否与第一个图像匹配。如果是，则返回 true，否则返回 false。
因此，如果输入类似于 n = 4，first = {"..x.", "x.x.", "x.xx", "xx.."}，second = {"..xx", "x.xx", ".x.x", "..x."}，则输出将为 False。
要解决此问题，我们将遵循以下步骤−
Define a function find(), this will take an array of pairs x, an array of pairs y,
d1 := first value of y[0] - first value of x[0]
d2 := second value of y[1] - second value of x[1]
for initialize i := 1, when i < size of x, update (increase i by 1), do:
if first value of y[i] - first value of x[i] is not equal to d1 or second value of y[i] - second value of x[i] is not equal to d2, then:
return false
return true
Define a function rotate(), this will take n, an array of pairs a, an array of pairs b,
for initialize i := 0, when i < size of b, update (increase i by 1), do:
b[i] := make_pair(second value of b[i], n - first value of b[i] - 1)
Define two arrays a, b that can contain integer pairs
for initialize i := 0, when i < n, update (increase i by 1), do:
s := first[i]
for initialize j := 0, when j < n, update (increase j by 1), do:
if s[j] is same as 'x', then:
insert pair(i, j) at the end of a
for initialize i := 0, when i < n, update (increase i by 1), do:
s := second[i]
for initialize j := 0, when j < n, update (increase j by 1), do:
if s[j] is same as 'x', then:
insert pair(i, j) at the end of b
if size of a is not equal to size of b, then:
return false
if size of a is same as 0, then:
return true
check := false
sort the array a
for initialize i := 0, when i < 4, update (increase i by 1), do:
sort the array b
if find(a, b), then:
check := true
rotate(n, a, b)
if check is true, then:
return true
Otherwise
return false示例
让我们看看以下实现，以更好地理解−
#include <bits/stdc++.h>
using namespace std;
bool find(vector<pair<int, int>> x, vector<pair<int, int>> y){
int d1 = y[0].first - x[0].first;
int d2 = y[1].second - x[1].second;
for(int i = 1; i < x.size(); i++){
if(y[i].first - x[i].first != d1 || y[i].second - x[i].second != d2){
return false;
}
}
return true;
}
void rotate(int n, vector<pair<int, int>> a, vector<pair<int, int>> b){
for(int i = 0; i < b.size(); i++){
b[i] = make_pair(b[i].second, n - b[i].first - 1);
}
}
bool solve(int n, vector<string> first, vector<string> second){
vector<pair<int, int>> a, b;
for(int i = 0; i < n; i++){
string s = first[i];
for(int j = 0; j < n; j++){
if(s[j] == 'x'){
a.push_back(make_pair(i, j));
}
}
}
for(int i = 0; i < n; i++){
string s = second[i];
for(int j = 0; j < n; j++){
if(s[j] == 'x'){ b.push_back(make_pair(i,j));
}
}
}
if(a.size() != b.size()){
return false;
}
if(a.size() == 0){
return true;
}
bool check = false;
sort(a.begin(),a.end());
for(int i = 0; i < 4; i++){
sort(b.begin(),b.end());
if(find(a,b)){
check = true;
}
rotate(n, a, b);
}
if(check){
return true;
}else{
return false;
}
}
int main() {
int n = 4; vector<string> first = {"..x.", "x.x.", "x.xx", "xx.."}, second = {"..xx", "x.xx", ".x.x", "..x."};
cout<< solve(n, first, second);
return 0;
}输入
4, {"..x.", "x.x.", "x.xx", "xx.."}, {"..xx", "x.xx", ".x.x", "..x."}
输出
0
更新日期：02-Mar-2022
190 次浏览
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