C ++程序检查旋转和平移后两个图像是否匹配
假设第一个和第二个有两个 n * n 像素的正方形图像。像素可以是黑色或白色。图像以矩阵表示形式给出，如果像素为黑色，则表示为“x”，如果为白色，则表示为“.”。我们必须在 90° 旋转和平移后检查第二张图像与第一张图像的匹配情况。如果是，我们返回真，否则，我们返回假。
因此，如果输入类似于 n = 4，则 first = {"..x.", "xx", "x.xx", "xx.."}, second = {"..xx", "x. xx”、“.xx”、“..x.”}，则输出将为 False。
为了解决这个问题，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."}输出结果0
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