CSES-solutions/CSES Problem Set/Rectangle Cutting.cpp
Segcolt 96022df10c Add a cses dp problem
Trying to learn more about range dp.
2024-09-27 20:50:37 -03:00

106 lines
2.1 KiB
C++

/* Problem URL: https://cses.fi/problemset/task/1744 */
#include <bits/stdc++.h>
using namespace std;
#define V vector
#define rmin(a, b) a = min(a, b)
#define rmax(a, b) a = max(a, b)
#define rep(i, lim) for (size_t i = 0; i < (lim); i++)
#define nrep(i, s, lim) for (size_t i = s; i < (lim); i++)
#define repv(i, v) for (auto &i : (v))
#define fillv(v) for (auto &itr_ : (v)) { cin >> itr_; }
#define sortv(v) sort(v.begin(), v.end())
#define all(v) (v).begin(), (v).end()
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vvvvi = vector<vvvi>;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vvvl = vector<vvl>;
using vvvvl = vector<vvvl>;
template<class v>
auto operator<<(ostream &os, const vector<v> &vec)->ostream& {
os << vec[0];
for (size_t i = 1; i < vec.size(); i++) {
os << ' ' << vec[i];
}
os << '\n';
return os;
}
template<class v>
auto operator>>(istream &is, vector<v> &vec)->istream& {
for (auto &i : vec) {
is >> i;
}
return is;
}
template<class v>
auto operator<<(ostream &os, const vector<vector<v>> &vec)->ostream& {
for (auto &i : vec) {
os << i[0];
for (size_t j = 1; j < i.size(); j++) {
os << ' ' << i[j];
}
os << '\n';
}
return os;
}
template<class v>
auto operator>>(istream &is, vector<vector<v>> &vec)->istream& {
for (auto &i : vec) {
for (auto &j : i) {
is >> j;
}
}
return is;
}
const ll inf = numeric_limits<ll>::max() >> 1;
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int a, b;
cin >> a >> b;
if (a > b) {
swap(a, b);
}
vvl dp(a + 1, vl(b + 1, inf));
for (size_t i = 0; i <= a; i++) {
dp[i][i] = 0;
}
for (size_t i = 1; i <= a; i++) {
for (size_t j = 1; j <= b; j++) {
for (size_t k = 1; k <= (b / 2); k++) {
if (k < i) {
dp[i][j] = min(dp[i][j], dp[k][j] + dp[i - k][j] + 1);
}
if (k < j) {
dp[i][j] = min(dp[i][j], dp[i][k] + dp[i][j - k] + 1);
}
}
}
}
cout << dp[a][b] << '\n';
}