感觉我dp的状态定义和其他的人都不一样。
dp[j][k]表示拥有j股还要等待k天的最大值，这样是可以把第i天这一维给省略掉的。（这样做的话要给间隔w++，转换成等待的时间）
状态转移：
每一天开始，都要把所有等待时间给减一天，f[j][k] = f[j][k+1]
接着就是卖出和买入
卖出：$f[j][w] = max(f[t][0] + (t - j) * b[i]) ，其中 j < t <= min(Maxp, j + d[i])$
买入：$f[j][w] = max(f[t][0] + (j - t) * a[i]) ，其中(max(0, j - c[i]) <= t < j)$

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 2010;

int n, m, p, a[N], b[N], c[N], d[N], f[N][N], z[N];

int main() {
    scanf("%d%d%d", &n, &m, &p); // m = number, p = day
    for (int i = 1; i <= n; i++)
        scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
    memset(f, 0xcf, sizeof(f)); f[0][0] = 0; 
    int inf = f[0][1], ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            f[j][0] = max(f[j][0], f[j][1]);
            for (int k = 1; k < p; k++) f[j][k] = f[j][k + 1];
            if (p >= 1) f[j][p] = inf;
        }
        for (int j = 0; j <= m; j++) z[j] = f[j][0];
        //buy
        for (int j = 1; j <= m; j++) 
            for (int k = 1; k <= min(c[i], j); k++)
                f[j][p] = max(f[j][p], z[j - k] - k * a[i]),
                ans = max(ans, f[j][p]);
        //sell  
        for (int j = 0; j < m; j++)
            for (int k = 1; j + k <= m && k <= d[i]; k++)
                f[j][p] = max(f[j][p], z[j + k] + k * b[i]),
                ans = max(ans, f[j][p]);
    }
    cout << ans << endl;
    return 0;
}

但是这样子维护是$O(N^3)$的，所以我们要优化状态转移
1.
$f[j][k] = f[j][k+1]$
我们发现只是让f往左边平移了一个，总体的顺序是不变的，所以我们只要将一个时间戳++
2.
$ f[t][0] + (t - j) * b[i] = f[t][0] + t * b[i] - j * b[i] $
$ (f[t][0] + t * b[i])就可以用单调队列维护了 $
时间变成$O(N^3)$

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 2010;

int n, m, p, a[N], b[N], c[N], d[N], f[N][N], z[N];

int l, r, q[N];

int main() {
    scanf("%d%d%d", &n, &m, &p); p++; 
    for (int i = 1; i <= n; i++)
        scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
    memset(f, 128, sizeof(f)); f[0][0] = 0; 
    int inf = f[0][1], ans = 0, time = 0;
    for (int i = 1; i <= n; i++) {
        int jn = time;
        time++; if (time == p + 1) time = 0;
        for (int j = 0; j <= m; j++) {
            f[j][time] = max(f[j][time], f[j][jn]);
            if (p >= 1) f[j][jn] = inf;
        }
        for (int j = 0; j <= m; j++) z[j] = f[j][time];
        //buy
        l = r = 1, q[1] = 0;
        for (int j = 1; j <= m; j++) {
            while (l <= r && q[l] < max(j - c[i], 0)) l++;
            f[j][jn] = max(f[j][jn], z[q[l]] - (j - q[l]) * a[i]);
            ans = max(ans, f[j][jn]);
            while (l <= r && z[q[r]] + q[r] * a[i] <= z[j] + j * a[i]) r--;
            q[++r] = j;
        }
        //sell  
        l = r = 1, q[1] = m;
        for (int j = m - 1; j >= 0; j--) {
            while (l <= r && q[l] > min(m, j + d[i])) l++;
            f[j][jn] = max(f[j][jn], z[q[l]] + (q[l] - j) * b[i]);
            ans = max(ans, f[j][jn]);
            while (l <= r && z[q[r]] + q[r] * b[i] <= z[j] + j * b[i]) r--;
            q[++r] = j;
        }
    }
    cout << ans << endl;
    return 0;
}