Codeforces 18D Seller Bob DP

Test instructions: In order you can get some value for 2^x hard drive, you can keep one and sell it later, but you can not keep other hard drives before selling, for maximum benefit

Idea: Dp[i]=min (Dp[i-1],dp[j]+2^x[j])

ImportJava.math.BigInteger;ImportJava.util.Scanner; Public classMain { Public StaticBigInteger MAX (BigInteger A,biginteger b) {if(A.compareto (b) ==1)returnA; returnb; }     Public Static voidMain (string[] args) {BigInteger dp[]=Newbiginteger[5005]; BigInteger d[]=Newbiginteger[2005];  for(inti=0;i<=2000;i++){            if(i==0) d[i]=Biginteger.one; ElseD[i]=d[i-1].multiply (Biginteger.valueof (2)); }        intf[]=New int[5005]; intp[]=New int[5005]; Scanner Cin=NewScanner (system.in);  while(Cin.hasnext ()) {intn=Cin.nextint ();  for(inti=0;i<n;i++) {String s=Cin.next (); intx=Cin.nextint (); if(S.charat (0) = = ' W ') f[i]=0; ElseF[i]=1; P[i]=x; }             for(inti=0;i<n;i++) Dp[i]=Biginteger.zero;  for(inti=1;i<n;i++){                if(f[i]==1){                     for(intj=i-1;j>=0;j--){                        if(f[j]==0&&p[i]==P[j]) {Dp[i]=max (dp[i-1], Dp[j].add (D[p[i]));  Break; } Dp[i]=max (Dp[i], dp[i-1]); }                }                ElseDp[i]=dp[i-1]; } System.out.println (Dp[n-1]); }    }}

Codeforces 18D Seller Bob DP