Submission #211919

#	Time	Username	Problem	Language	Result	Execution time	Memory
211919		mode149256	Raisins (IOI09_raisins)	C++14	100 / 100	323 ms	33784 KiB

raisins

/*input
2 3
2 7 5
1 9 5
*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

ll dp[50][50][50][50];

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N, M;
	cin >> N >> M;
	vii sk(N, vi(M));
	vii sm(N, vi(M, 0));

	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			cin >> sk[i][j];
			sm[i][j] = sk[i][j];
			if (i) sm[i][j] += sm[i - 1][j];
			if (j) sm[i][j] += sm[i][j - 1];
			if (i and j) sm[i][j] -= sm[i - 1][j - 1];
		}
	}

	auto sum = [&](int ln, int lm, int hn, int hm) {
		return sm[hn][hm]
		       - (lm ? sm[hn][lm - 1] : 0)
		       - (ln ? sm[ln - 1][hm] : 0)
		       + (lm and ln ? sm[ln - 1][lm - 1] : 0);
	};



	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			for (int n = 0; n + i < N; ++n)
			{
				for (int m = 0; m + j < M; ++m)
				{
					if (!i and !j) {
						dp[n][m][n + i][m + j] = 0;
						continue;
					}
					dp[n][m][n + i][m + j] = (ll)1e16;
					for (int ki = 0; ki < i; ++ki)
					{
						ll nauja = dp[n][m][n + ki][m + j]
						           + dp[n + ki + 1][m][n + i][m + j];

						dp[n][m][n + i][m + j] = min(dp[n][m][n + i][m + j], nauja);
					}

					for (int kj = 0; kj < j; ++kj)
					{
						ll nauja = dp[n][m][n + i][m + kj]
						           + dp[n][m + kj + 1][n + i][m + j];

						dp[n][m][n + i][m + j] = min(dp[n][m][n + i][m + j], nauja);
					}
					dp[n][m][n + i][m + j] += sum(n, m, n + i, m + j);
				}
			}
		}
	}
	printf("%lld\n", dp[0][0][N - 1][M - 1]);
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

• Subtask 1: 100 / 100