AcWing

https://www.acwing.com/problem/content/334/

第一次写单调队列优化的dp，首先朴素的做法不难想到，就是复杂度 (O(n^3)) ，然后考虑优化。

每天都从 (pre=max(0,i-w-1)) 天转移过来就刚刚好了。

考虑每个k是怎么更新j的。

买入股票：

(dp[i][j]=max{dp[pre][k]-(j-k)*AP_i;|;k leq j;and;(j-k) leq AS_i})

(dp[i][j]=max{dp[pre][k]+k*AP_i;|;k leq j;and;(j-k) leq AS_i}-j*AP_i)

里面的这个k的取值范围随j的增长会滑动，所以用单调队列维护。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll INF = 0x3f3f3f3f3f3f3f3f;
ll dp[2005][2005];

deque<int> AQ, BQ;

int AP, BP;

ll Calc_A(int id1, int id2) {
    return dp[id1][id2] + 1ll * id2 * AP;
}
ll Calc_B(int id1, int id2) {
    return dp[id1][id2] + 1ll * id2 * BP;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int T, maxP, W;
    scanf("%d%d%d", &T, &maxP, &W);
    memset(dp, -INF, sizeof(dp));
    dp[0][0] = 0;
    for(int i = 1; i <= T; ++i) {
        for(int j = 0; j <= maxP; ++j)
            dp[i][j] = dp[i - 1][j];
        int AS, BS;
        scanf("%d%d%d%d", &AP, &BP, &AS, &BS);

        AQ.clear(), BQ.clear();
        int pre = max(0, i - W - 1);
        int Ak = 0;
        for(int j = 0; j <= maxP; ++j) {
            while(Ak <= j) {
                ll tmp = Calc_A(pre, Ak);
                while(!AQ.empty() && Calc_A(pre, AQ.back()) <= tmp)
                    AQ.pop_back();
                AQ.push_back(Ak);
                ++Ak;
            }
            while(!AQ.empty() && (AQ.front() < j - AS))
                AQ.pop_front();
            if(!AQ.empty())
                dp[i][j] = max(dp[i][j], Calc_A(pre, AQ.front()) - 1ll * j * AP);
            //cout<<AQ.size()<<endl;
        }

        int Bk = maxP;
        for(int j = maxP; j >= 0; --j) {
            while(Bk >= j) {
                ll tmp = Calc_B(pre, Bk);
                while(!BQ.empty() && Calc_B(pre, BQ.back()) <= tmp)
                    BQ.pop_back();
                BQ.push_back(Bk);
                --Bk;
            }
            while(!BQ.empty() && (BQ.front() > BS + j))
                BQ.pop_front();
            if(!BQ.empty())
                dp[i][j] = max(dp[i][j], Calc_B(pre, BQ.front()) - 1ll * j * BP);
            //cout<<BQ.size()<<endl;
        }

        /*for(int j = 0; j <= maxP; ++j) {
            printf("dp[%d][%d]=%lld
", i, j, dp[i][j]);
        }
        puts("");*/
    }
    ll ans = -INF;
    for(int i = 0; i <= maxP; ++i) {
        ans = max(ans, dp[T][i]);
    }
    printf("%lld
", ans);
}