【SCOI2010】股票交易

Description

【SCOI2010】股票交易

在T天时间内，第\(i\)天股票购入价为\(ap_i\)，出售价为\(bp_i\)，每天最多购入\(as_i\)股，最多出售\(bs_i\)股

任意时刻手中的股票数不能超过\(Maxp\)，且两次交易至少间隔\(W\)天

最大化收益，初始资金视为无限大

Solution

单调队列+dp

根据题意不难设计状态，定义\(f[i][j]\)表示前\(i\)天，手里最后有\(j\)个股票的最大收益，那么状态转移方程就是：

\[f[i][j]= \begin{cases} f[i-1][j] \\ f[i-w-1][k]-ap[i]*(j-k) & k\in[j-as[i],j)\\ f[i-w-1][k]+bp[i]*(k-j) & k\in(j,j+bs[i]] \end{cases} \]

以第二个式子为例，我们可以变形得到\(f[i][j]=f[i-w-1][k]+ap[i]*k-ap[i]*j\)

这个式子符合单调队列的一般形式，所以我们可以用单调队列进行优化

第三个式子同理

时间复杂度为\(O(TMaxp)\)

Code

#include <bits/stdc++.h>
// check if it is judged online
namespace shl {
	typedef long long ll;
	inline int read() {
		int ret = 0, op = 1;
		char c = getchar();
		while (!isdigit(c)) {
			if (c == '-') op = -1;
			c = getchar();
		}
		while (isdigit(c)) {
			ret = ret * 10 + c - '0';
			c = getchar();
		}
		return ret * op;
	}
	const int N = 2010;
	int n, m, w;
	int as[N], bs[N], ap[N], bp[N];
	int head, tail, que[N];
	int f[N][N];
	inline int calca(int i, int ww, int k) {
		return f[i - ww - 1][k] + ap[i] * k;
	}
	inline int calcb(int i, int ww, int k) {
		return f[i - ww - 1][k] + bp[i] * k;
	}
	using std::max;
	int main() {
		n = read(), m = read(), w = read();
		for (register int i = 1; i <= n; ++i) {
			ap[i] = read(), bp[i] = read(), as[i] = read(), bs[i] = read();
		} 
		memset(f, -0x7f, sizeof(f));
		for (register int i = 0; i <= n; ++i)  f[i][0] = 0;
		for (register int i = 1; i <= n; ++i) {
			for (register int j = 0; j <= as[i]; ++j) f[i][j] = -ap[i] * j;
			for (register int j = m; j >= 0; --j) f[i][j] = max(f[i][j], f[i - 1][j]);
			if (i - w - 1 < 0) continue ;
			head = 1, tail = 0;
			for (register int j = 0; j <= m; ++j) {
				while (head <= tail && que[head] < j - as[i]) head++;
				while (head <= tail && calca(i, w, j) >= calca(i, w, que[tail])) tail--;
				que[++tail] = j;
				if (head <= tail) f[i][j] = max(f[i][j], calca(i, w, que[head]) - j * ap[i]);
			}
			head = 1, tail = 0;
			for (register int j = m; j >= 0; --j) {
				while (head <= tail && que[head] > j + bs[i]) head++;
				while (head <= tail && calcb(i, w, j) >= calcb(i, w, que[tail])) tail--;
				que[++tail] = j;
				if (head <= tail) f[i][j] = max(f[i][j], calcb(i, w, que[head]) - j * bp[i]);
			}			
		}	
		int ans = 0;	
		for (register int i = 0; i <= m; ++i) ans = max(ans, f[n][i]);
		printf("%d\n", ans);
		return 0;
	}
}
int main() {
#ifdef LOCAL
	freopen("textname.in", "r", stdin);
	freopen("textname.out", "w", stdout);
#endif
	shl::main();
	return 0;
}