P5858 「SWTR-03」Golden Sword

思路1

$f_{i,j}$表示放入$i$原料，并且当前锅中有$j$个原料。
状态转移方程：$f_{i,j}=\underset{j-1\le k\le min\{j+s-1\}}{max}\{f_{i-1,k}+a_i\times j\}$

代码1

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 5010;
int n,w,s;
LL a[N];
LL f[N][N];
int main () {
	cin >> n >> w >> s;
	for (int i = 1;i <= n;i++) cin >> a[i];
	memset (f,-0x3f,sizeof (f));
	f[0][0] = 0;
	for (int i = 1;i <= n;i++) {
		for (int j = 1;j <= w;j++) {
			for (int k = j - 1;k <= min (w,j + s - 1);k++) f[i][j] = max (f[i][j],f[i - 1][k] + j * a[i]);
		}
	}
	LL ans = -1e18;
	for (int i = 0;i <= w;i++) ans = max (ans,f[n][i]);
//	for (int i = 1;i <= n;i++) {
//		for (int j = 0;j <= w;j++) cout << f[i][j] << ' ';
//		cout << endl;
//	}
//	cout << endl;
	cout << ans << endl;
	return 0;
}

思路2

公式一拆变为$f_{i,j}=\underset{j-1\le k\le min\{j+s-1\}}{max}\{f_{i-1,k}\}+a_i\times j$
我们用单调队列$O(1)$查询f_{i - 1,k}的最大值。

代码2

#include <iostream>
#include <cstring>
#include <deque>
using namespace std;
typedef long long LL;
const int N = 5010;
int n,w,s;
LL a[N];
LL f[N][N];
int main () {
	cin >> n >> w >> s;
	for (int i = 1;i <= n;i++) cin >> a[i];
	memset (f,-0x3f,sizeof (f));
	f[0][0] = 0;
	for (int i = 1;i <= n;i++) {
		deque <int> q;
		q.push_back (w);
		for (int j = w;j >= 1;j--) {
			while (q.size () && q.front () - j >= s) q.pop_front ();
			while (q.size () && f[i - 1][q.back ()] <= f[i - 1][j - 1]) q.pop_back ();
			q.push_back (j - 1);
			f[i][j] = f[i - 1][q.front ()] + a[i] * j;
		}
	}
	LL ans = -1e18;
	for (int i = 0;i <= w;i++) ans = max (ans,f[n][i]);
//	for (int i = 1;i <= n;i++) {
//		for (int j = 0;j <= w;j++) cout << f[i][j] << ' ';
//		cout << endl;
//	}
//	cout << endl;
	cout << ans << endl;
	return 0;
}