Created
February 3, 2019 13:44
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#include <iostream> | |
#include <string> | |
#include <vector> | |
using namespace std; | |
// Вычисляем максимальный квадрат с правым нижним | |
// углом в точке (i,j) на основе ужн вычесленного для точки | |
// (i-1,j-1) . Для этого нам также надо знать сколько | |
// свободных ячеек левее и выше точки (i,j) | |
std::pair<int, int> Solve(const vector<string>& a) | |
{ | |
vector<vector<int>> r(a.size(),vector<int>(a[0].size(), -1)); | |
vector<int> v(a[0].size(), -1); | |
int m = 0; | |
int count = 0; | |
for (int i = 0; i < a.size(); ++i) { | |
int h = -1; | |
for (int j = 0; j < a[0].size(); ++j) { | |
if (a[i][j] == '1' || j == 0 || i == 0) { | |
auto c = a[i][j] != '1'; | |
r[i][j] = c; | |
h = c ? h : j; | |
v[j] = c ? v[j] : i; | |
continue; | |
} | |
auto _a = r[i-1][j-1]+1; | |
auto _b = i - v[j]; | |
auto _c = j - h; | |
r[i][j] = min(_a, min(_b, _c)); | |
if (r[i][j] == m) { | |
++count; | |
} | |
if (m < r[i][j]) { | |
m = r[i][j]; | |
count = 1; | |
} | |
} | |
} | |
return make_pair(m, count); | |
} | |
int main() { | |
{ | |
auto r = Solve( | |
{ | |
"100001", | |
"010000", | |
"100000", | |
"001000", | |
"000010", | |
"001000" | |
}); | |
cout << r.first << " " << r.second << endl; | |
} | |
{ | |
auto r = Solve( | |
{ | |
"111111", | |
"111111", | |
"111111", | |
"111111", | |
"111111", | |
"111111" | |
}); | |
cout << r.first << " " << r.second << endl; | |
} | |
{ | |
auto r = Solve( | |
{ | |
"000000", | |
"000000", | |
"000000", | |
"000000", | |
"000000", | |
"000000" | |
}); | |
cout << r.first << " " << r.second << endl; | |
} | |
return 0; | |
} |
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