Code Forces 18D Seller Bob(简单DP)

D. Seller Bob

time limit per test

2 seconds

memory limit per test

128 megabytes

input

standard input

output

standard output

Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:

• A customer came to Bob and asked to sell him a 2x MB
  memory stick. If Bob had such a stick, he sold it and got 2x berllars.
• Bob won some programming competition and got a 2x MB
  memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.

Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing
all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much
money he could have earned, if he had acted optimally.

Input

The first input line contains number n (1 ≤ n ≤ 5000)
— amount of Bob's working days. The following n lines contain the description of the days. Line sell
x stands for a day when a customer came to Bob to buy a 2x MB
memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there
is not more than one line sell x. Line win x stands for
a day when Bob won a 2x MB
memory stick (0 ≤ x ≤ 2000).

Output

Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time.

Examples

input

7
win 10
win 5
win 3
sell 5
sell 3
win 10
sell 10

output

1056

input

3
win 5
sell 6
sell 4

output

0

思路：动态规划，用一个数组表示二进制数，来表示可以赚的钱，这里可以用stl里面的bitset，最后一次性输出，当然也可以用高精度

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
#include <vector>
#include <bitset>

using namespace std;
int n;
int pre[2005];
int a[5005];
int ans[5005];
vector<int> s;
bitset <2005> dp[5005],y;
string b;
int main()
{
   scanf("%d",&n);
   memset(pre,0,sizeof(pre));
   s.clear();

   dp[0]=0;
   for(int i=1;i<=n;i++)
   {
       cin>>b>>a[i];
       if(b[0]=='w')
       {pre[a[i]]=i;dp[i]=dp[i-1];}
       else
       {
           if(!pre[a[i]]) {dp[i]=dp[i-1];continue;}
           y=dp[pre[a[i]]];
           y[a[i]]=1;
           for(int j=2000;j>=0;j--)
           {
               if(dp[i-1][j]>y[j]){dp[i]=dp[i-1];break;}
               if(dp[i-1][j]<y[j]){dp[i]=y;break;}
               if(j==0){dp[i]=y;}
           }
       }
   }
   s.push_back(0);
   for(int i=dp[n].size()-1;i>=0;i--)
   {
       int k=0;
       for(int j=0;j<s.size();j++)
       {
           int now=s[j];
           s[j]=(now*2+k)%10;
           k=(now*2+k)/10;
       }
       if(k)
        s.push_back(k);
       if(dp[n][i])
       {
           int k=0;s[0]++;
           for(int j=0;j<s.size();j++)
           {
               int now=s[j];
               s[j]=(now+k)%10;
               k=(now+k)/10;
           }
           if(k)
            s.push_back(k);
       }
   }
   for(int i=s.size()-1;i>=0;i--)
    printf("%d",s[i]);
   cout<<endl;
   return 0;

}