zoukankan      html  css  js  c++  java
  • 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;
    
    }



  • 相关阅读:
    路径规划算法总结
    常用滤波器整理
    Debian 9 strech 安装 ROS lunar
    understand 安装笔记
    protobuf 安装与卸载
    maven-surefire-plugin
    spring数据源、数据库连接池
    日志插件总结
    pom.xml常用元素解析
    BeanFactory笔记
  • 原文地址:https://www.cnblogs.com/dacc123/p/8228742.html
Copyright © 2011-2022 走看看