HDU-3401 Trade 单调队列优化DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3401

　　DP方程容易想出来，f[i][j]表示第i天拥有j个股票的最优解，则：

　　　　1、不买不卖，f[i][j]=Max{ f[i][j], f[i-1][j] }。

　　　　2、买进，f[i][j]=Max{ f[i][j], f[pre][k] - (j-k)*ap[i] | j>=k }。

　　　　3、卖出，f[i][j]=Max{ f[i][j], f[pre][k] +(k-j)*bp[i] | k>=j }。

　　直接转移复杂度O(n^3)，超时。考虑第二种情况，f[pre][k] - (j-k)*ap[i] = (f[pre][k]+k*ap[i])-j*ap[i]，用单调队列维护这个值(f[pre][k]+k*ap[i])就可以了。情况3类似。

//STATUS:C++_AC_375MS_16132KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=2010;
const int INF=0x3f3f3f3f;
const int MOD=1e+7,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End
int f[N][N],ap[N],bp[N],as[N],bs[N],q[N*2][2];
int T,n,m,d;
int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,k,hig,ans,front,rear,p;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&d);
        for(i=1;i<=n;i++){
            scanf("%d%d%d%d",&ap[i],&bp[i],&as[i],&bs[i]);
        }
        mem(f,-INF);
        for(i=1;i<=d+1;i++){
            for(j=0;j<=as[i];j++)
                f[i][j]=-j*ap[i];
        }
        for(i=1;i<=d+1;i++){
            for(j=0;j<=m;j++)
                f[i][j]=max(f[i][j],f[i-1][j]);
        }
        for(i=d+2;i<=n;i++){
            p=i-d-1;
            for(j=0;j<=m;j++)f[i][j]=f[i-1][j];
            //
            front=rear=0;
            q[rear][0]=f[p][0];q[rear++][1]=0;
            for(j=1;j<=m;j++){
                while(front<rear && q[front][1]<j-as[i])front++;
                f[i][j]=max(f[i][j],q[front][0]-j*ap[i]);
                while(front<rear && q[rear-1][0]<=f[p][j]+j*ap[i])rear--;
                q[rear][0]=f[p][j]+j*ap[i];q[rear++][1]=j;
            }
            //
            front=rear=0;
            q[rear][0]=f[p][m]+m*bp[i];q[rear++][1]=m;
            for(j=m-1;j>=0;j--){
                while(  front<rear && q[front][1]>j+bs[i])front++;
                f[i][j]=max(f[i][j],q[front][0]-j*bp[i]);
                while(front<rear && q[rear-1][0]<=f[p][j]+j*bp[i])rear--;
                q[rear][0]=f[p][j]+j*bp[i];q[rear++][1]=j;
            }
        }
        ans=0;
        for(i=0;i<=m;i++)ans=max(ans,f[n][i]);
        printf("%d
",ans);
    }
    return 0;
}