【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 
// 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;
}