제출 #125888

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
125888		tutis	조교 (CEOI16_popeala)	C++17	17 / 100	2066 ms	5996 KiB

popeala

/*input
2 3 3
4 3 5
101
110
*/
#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const ll modulo = 1000000007;
struct line
{
	ll k;
	ll b;
	// kx + b
};
struct ST
{
	int l, r;
	vector<line>K;
	ST *left;
	ST *right;
	ll ans[51];
	ST(int l, int r): l(l), r(r)
	{
		if (l < r)
		{
			left = new ST(l, (l + r) / 2);
			right = new ST((l + r) / 2 + 1, r);
		}
	}
	void fix(const vector<line>a)
	{
		if (l < r)
		{
			left->fix(a);
			right->fix(a);
			for (int x = 0; x <= 50; x++)
				ans[x] = min(left->ans[x], right->ans[x]);
		}
		else
		{
			for (int x = 0; x <= 50; x++)
			{
				ans[x] = a[l].k * x + a[l].b;
			}
		}
	}
	ll mini(int x, int y, ll xx)
	{
		if (x <= l && r <= y)
			return ans[xx];
		if (r < x || y < l)
			return 1e15;
		return min(left->mini(x, y, xx), right->mini(x, y, xx));
	}
};
int main()
{
	ios_base::sync_with_stdio(false);
	ll N, T, S;
	cin >> N >> T >> S;
	ll p[T + 1];
	p[0] = 0;
	for (ll i = 1; i <= T; i++)
	{
		cin >> p[i];
		p[i] += p[i - 1];
	}
	string res[N];
	for (ll i = 0; i < N; i++)
		cin >> res[i];
	ll dp[S + 1][T + 1];
	for (ll a = 0; a <= S; a++)
	{
		for (ll b = 0; b <= T; b++)
		{
			dp[a][b] = 1e11;
		}
	}
	dp[0][0] = 0;
	int kada[T + 1][N + 1];
	for (int i = 0; i <= N; i++)
		kada[0][i] = 0;
	for (ll b = 1; b <= T; b++)
	{
		for (ll i = 0; i < N; i++)
		{
			if (res[i][b - 1] == '0')
			{
				kada[b][i] = b;
			}
			else
				kada[b][i] = kada[b - 1][i];
		}
		kada[b][N] = b;
	}
	for (ll b = 1; b <= T; b++)
		sort(kada[b], kada[b] + N + 1);
	ST medis(0, T);
	for (int c = 1; c <= S; c++)
	{
		vector<line>aaa;
		for (int a = 0; a <= T; a++)
		{
			aaa.push_back(line());
			aaa.back().b = dp[c - 1][a];
			aaa.back().k = -p[a];
		}
		medis.fix(aaa);
		for (int b = 1; b <= T; b++)
		{
			int l = 1;
			for (int x = 0; x <= N; x++)
			{
				int r = kada[b][x];
				ll mini = medis.mini(l - 1, r - 1, x);
				dp[c][b] = min(dp[c][b], mini + x * p[b]);
				l = r + 1;
			}
		}
	}
	for (ll c = 1; c <= S; c++)
	{
		cout << dp[c][T] << "\n";
	}
}

• Subtask 1: 8 / 8
• Subtask 2: 9 / 9
• Subtask 3: 0 / 9
• Subtask 4: 0 / 74