BZOJ1855或洛谷2569 [SCOI2010]股票交易

一道单调队列优化\(DP\)

BZOJ原题链接

洛谷原题链接

朴素的\(DP\)方程并不难想。

定义\(f[i][j]\)表示到第\(i\)天，手上持有\(j\)股时的最大收益。

转移方程可以分成四个部分。

1. 第\(i\)天为空手时买股票

\(\qquad\qquad f[i][j]=-AP_i\times j\)

1. 第\(i\)天不进行交易

\(\qquad\qquad f[i][j]=\max\{f[i][j],f[i-1][j]\}\)

1. 第\(i\)天在之前基础上买

\(\qquad\qquad f[i][j]=\max\limits_{k=\max\{j-AS_i,1\}}^{j-1}\{f[i][j],f[i-w-1][k]-AP_i\times (j-k)\}\)

1. 第\(i\)天在之前基础上卖

\(\qquad\qquad f[i][j]=\max\limits_{k=j+1}^{\min\{j+BS_i,MaxP\}}\{f[i][j],f[i-w-1][k]+BP_i\times (k-j)\}\)

显然决策增减满足单调性，可以用单调队列维护，注意第\(4\)项要倒序才能维护。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 2010;
int f[N][N], AP[N], BP[N], AS[N], BS[N], q[N];
int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c<'0' || c>'9'; c = getchar())
		p = (c == '-' || p) ? 1 : 0;
	for (; c >= '0'&&c <= '9'; c = getchar())
		x = x * 10 + (c - '0');
	return p ? -x : x;
}
inline int minn(int x, int y)
{
	return x < y ? x : y;
}
inline int maxn(int x, int y)
{
	return x > y ? x : y;
}
int main()
{
	int i, j, n, m, w, l, r;
	n = re();
	m = re();
	w = re();
	memset(f, 250, sizeof(f));
	for (i = 1; i <= n; i++)
	{
		AP[i] = re();
		BP[i] = re();
		AS[i] = re();
		BS[i] = re();
		f[i][0] = 0;
	}
	for (i = 1; i <= n; i++)
	{
		for (j = 0; j <= AS[i]; j++)
			f[i][j] = -AP[i] * j;
		for (j = m; j >= 0; j--)
			f[i][j] = maxn(f[i][j], f[i - 1][j]);
		if (i - w - 1 > 0)
		{
			for (r = j = 0, l = 1; j <= m; j++)
			{
				while (l <= r && q[l] < j - AS[i])
					l++;
				while (l <= r && f[i - w - 1][j] + AP[i] * j >= f[i - w - 1][q[r]] + AP[i] * q[r])
					r--;
				q[++r] = j;
				if (l <= r)
					f[i][j] = maxn(f[i][j], f[i - w - 1][q[l]] - AP[i] * (j - q[l]));
			}
			for (r = 0, l = 1, j = m; j >= 0; j--)
			{
				while (l <= r && q[l] > j + BS[i])
					l++;
				while (l <= r && f[i - w - 1][j] + BP[i] * j >= f[i - w - 1][q[r]] + BP[i] * q[r])
					r--;
				q[++r] = j;
				if (l <= r)
					f[i][j] = maxn(f[i][j], f[i - w - 1][q[l]] + BP[i] * (q[l] - j));
			}
		}
	}
	printf("%d", f[n][0]);
	return 0;
}