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=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e+,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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*][];
 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=;i<=n;i++){
             scanf("%d%d%d%d",&ap[i],&bp[i],&as[i],&bs[i]);
         }
         mem(f,-INF);
         for(i=;i<=d+;i++){
             for(j=;j<=as[i];j++)
                 f[i][j]=-j*ap[i];
         }
         for(i=;i<=d+;i++){
             for(j=;j<=m;j++)
                 f[i][j]=max(f[i][j],f[i-][j]);
         }
         for(i=d+;i<=n;i++){
             p=i-d-;
             for(j=;j<=m;j++)f[i][j]=f[i-][j];
             //
             front=rear=;
             q[rear][]=f[p][];q[rear++][]=;
             for(j=;j<=m;j++){
                 while(front<rear && q[front][]<j-as[i])front++;
                 f[i][j]=max(f[i][j],q[front][]-j*ap[i]);
                 while(front<rear && q[rear-][]<=f[p][j]+j*ap[i])rear--;
                 q[rear][]=f[p][j]+j*ap[i];q[rear++][]=j;
             }
             //
             front=rear=;
             q[rear][]=f[p][m]+m*bp[i];q[rear++][]=m;
             for(j=m-;j>=;j--){
                 while(  front<rear && q[front][]>j+bs[i])front++;
                 f[i][j]=max(f[i][j],q[front][]-j*bp[i]);
                 while(front<rear && q[rear-][]<=f[p][j]+j*bp[i])rear--;
                 q[rear][]=f[p][j]+j*bp[i];q[rear++][]=j;
             }
         }
         ans=;
         for(i=;i<=m;i++)ans=max(ans,f[n][i]);
         printf("%d\n",ans);
     }
     return ;
 }