bzoj1855: [Scoi2010]股票交易--单调队列优化DP

单调队列优化DP的模板题

不难列出DP方程：

对于买入的情况

由于dp[i][j]=max{dp[i-w-1][k]+k*Ap[i]-j*Ap[i]}

AP[i]*j是固定的，在队列中维护dp[i-w-1][k]+k*Ap[i]的单调性即可

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn = 2010;
 6 int T,maxP,w,Ap[maxn],Bp[maxn],As[maxn],Bs[maxn],dp[maxn][maxn],q[maxn*10];
 7
 8 int main(){
 9     scanf("%d%d%d", &T, &maxP, &w);
10     for (int i=1; i<=T; i++)
11         scanf("%d%d%d%d", &Ap[i], &Bp[i], &As[i], &Bs[i]);
12     memset(dp,-127,sizeof(dp));
13     for (int i=1; i<=T; i++){
14         for (int j=0; j<=As[i]; j++) dp[i][j]=-Ap[i]*j;
15         for (int j=0; j<=maxP; j++) dp[i][j]=max(dp[i][j], dp[i-1][j]);
16         if (i-w-1>=0){
17             int head=0, tail=0;
18             for (int j=0; j<=maxP; j++){
19                 while (head<tail && q[head]+As[i]<j) head++;
20                 while (head<tail && dp[i-w-1][j]+j*Ap[i] >= dp[i-w-1][q[tail-1]]+q[tail-1]*Ap[i]) tail--;
21                 q[tail++]=j;
22                 if (head<tail) dp[i][j]=max(dp[i][j], dp[i-w-1][q[head]]-(j-q[head])*Ap[i]);
23             }
24             head=0, tail=0;
25             for (int j=maxP; j>=0; j--){
26                 while (head<tail && q[head]-Bs[i]>j) head++;
27                 while (head<tail && dp[i-w-1][j]+j*Bp[i] >= dp[i-w-1][q[tail-1]]+q[tail-1]*Bp[i]) tail--;
28                 q[tail++]=j;
29                 if (head<tail) dp[i][j]=max(dp[i][j], dp[i-w-1][q[head]]+(q[head]-j)*Bp[i]);
30             }
31         }
32     }
33     int ans=0;
34     for (int i=0; i<=maxP; i++)
35         ans=max(ans, dp[T][i]);
36     printf("%d\n", ans);
37     return 0;
38 }