P2569 [SCOI2010]股票交易

${color{Pink}{>>Question}}$

令$f[i,j]$表示$i$天有$j$支股票的最大收益

有四种决策:不买不卖,不在之前的基础上买,买,卖

有

$$f[i,j] = maxegin{cases}
f[i-1,j]\
-ap_i*j\
underset{max(j-as_i,0) leq k < j}{max}left { f[i-w-1,k]+ap_i*k ight }-ap_i*j\
underset{j < k leq min(bs_i+j,MaxP)}{max}left { f[i-w-1,k]+bp_i*k ight }-bp_i*j
end{cases}$$

三四式就可以单掉队列优化

时间复杂度 $O(n*m)$

空间复杂度 $O(n*m)$

代码如下

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define ll long long
using namespace std;

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {x = 10*x + ch - '0'; ch = getchar();}
    x *= f;
}
//------------------------------------------------------------

const int N = 2005;

int n,m,w;
int l,r,q[N],as[N],bs[N];
ll ap[N],bp[N],f[N][N];

ll calc(int i,int k,int op) {
    return f[i-w-1][k]+ (op == 1 ? ap[i] : bp[i])*k;
}

int main() {
    int i,j; in(n); in(m); in(w);
    for(i = 1;i <= n; ++i) {
        in(ap[i]); in(bp[i]); in(as[i]); in(bs[i]);
    }
    memset(f,~0x3f,sizeof(f));
    f[0][0] = 0;
    for(i = 1;i <= n; ++i) {
        for(j = 0;j <= as[i]; ++j) f[i][j] = -ap[i]*j;
        for(j = 0;j <= m; ++j) f[i][j] = max(f[i][j],f[i-1][j]);
        if(i-w-1 < 0) continue;
        l = 1;r = 0;
        for(j = 0;j <= m; ++j) {
            while(l <= r && q[l] < max(0,j-as[i])) ++l;
            if(l <= r) f[i][j] = max(f[i][j],calc(i,q[l],1)-ap[i]*j);
            while(l <= r && calc(i,q[r],1) <= calc(i,j,1)) --r;
            q[++r] = j;
        }
        l = 1,r = 0;
        for(j = m;j >= 0; --j) {
            while(l <= r && q[l] > min(m,bs[i]+j)) ++l;
            if(l <= r) f[i][j] = max(f[i][j],calc(i,q[l],2)-bp[i]*j);
            while(l <= r&& calc(i,q[r],2) <= calc(i,j,2)) --r;
            q[++r] = j;
        }
    }
    ll ans = 0;
    for(i = 0;i <= m; ++i) ans = max(ans,f[n][i]);
    cout << ans;
    return 0;
}