zoukankan      html  css  js  c++  java
  • HDOJ 1087 Super Jumping! Jumping! Jumping!

    d( i )=d( j )『符合条件的0~i-1之间的最大的』+a[ i ]

    Super Jumping! Jumping! Jumping!

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 16601    Accepted Submission(s): 7066


    Problem Description
    Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now.



    The game can be played by two or more than two players. It consists of a chessboard(棋盘)and some chessmen(棋子), and all chessmen are marked by a positive integer or “start” or “end”. The player starts from start-point and must jumps into end-point finally. In the course of jumping, the player will visit the chessmen in the path, but everyone must jumps from one chessman to another absolutely bigger (you can assume start-point is a minimum and end-point is a maximum.). And all players cannot go backwards. One jumping can go from a chessman to next, also can go across many chessmen, and even you can straightly get to end-point from start-point. Of course you get zero point in this situation. A player is a winner if and only if he can get a bigger score according to his jumping solution. Note that your score comes from the sum of value on the chessmen in you jumping path.
    Your task is to output the maximum value according to the given chessmen list.
     


    Input
    Input contains multiple test cases. Each test case is described in a line as follow:
    N value_1 value_2 …value_N 
    It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
    A test case starting with 0 terminates the input and this test case is not to be processed.
     


    Output
    For each case, print the maximum according to rules, and one line one case.
     


    Sample Input
    3 1 3 24 1 2 3 44 3 3 2 10
     


    Sample Output
    4103
     


    Author
    lcy
     
     1 #include <iostream>
     2 #include <cstring>
     3 
     4 using namespace std;
     5 
     6 int dp[1111];
     7 int a[1111];
     8 
     9 int main()
    10 {
    11     int n;
    12     while(cin>>n&&n)
    13     {
    14         memset(dp,0,sizeof(dp));
    15         for(int i=1;i<=n;i++)
    16             cin>>a[i];
    17 
    18         for(int i=1;i<=n;i++)
    19         {
    20             int maxn=-999;
    21             for(int j=0;j<i;j++)
    22             {
    23                 if(a[i]>a[j])
    24                 maxn=max(maxn,dp[j]);
    25             }
    26             dp[i]=maxn+a[i];
    27         }
    28 
    29         int ans=-99999999;
    30         for(int i=0;i<=n;i++)
    31             ans=max(ans,dp[i]);
    32         cout<<ans<<endl;
    33     }
    34 
    35     return 0;
    36 }
  • 相关阅读:
    [转] java代码块 介绍
    win10多用户远程登录
    [读书笔记] learn python the hard way书中 有关powershell 的一些小问题
    Ansible 汇总
    Shell 研究
    Linux CenOS Python3 和 python2 共存
    Ansible安装
    MySQL 5.7.20绿色版安装详细图文教程
    Mysql常用运算符与函数汇总
    mysql 从陌生到熟练之----数据库备份恢复的实现方法
  • 原文地址:https://www.cnblogs.com/CKboss/p/3095975.html
Copyright © 2011-2022 走看看