• Super Jumping! Jumping! Jumping! ---HDU


    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.

    InputInput 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.
    OutputFor each case, print the maximum according to rules, and one line one case.
    Sample Input

    3 1 3 2
    4 1 2 3 4
    4 3 3 2 1
    0

    Sample Output

    4
    10
    3

    分析:LIS的变形处理,,dp
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<algorithm>
     5 using namespace std;
     6 const int maxn = 1010;
     7 int n;
     8 int a[maxn];
     9 int dp[maxn];
    10 
    11 
    12 int main(){
    13     while(~scanf("%d",&n)&&n){
    14         memset(a,0,sizeof(a));
    15         for( int i=1; i<=n; i++ ){
    16             scanf("%d",a+i);
    17         }
    18         int res=-1;
    19         for( int i=1; i<=n; i++ ){
    20             dp[i]=a[i];
    21             for( int j=1; j<i; j++ ){
    22                 if(a[i]>a[j]) dp[i]=max(dp[i],dp[j]+a[i]);
    23             }
    24             res=max(res,dp[i]);
    25         }
    26         cout<<res<<endl;
    27     }
    28     return 0;
    29 }
    有些目标看似很遥远,但只要付出足够多的努力,这一切总有可能实现!
  • 相关阅读:
    NYOJ 32(组合数)
    NYOJ 289(01背包)
    批量修改文件(图片)名称
    解决IIS7虚拟目录出现HTTP 错误 500.19(由于权限不足而无法读取配置文件)的问题
    MPP(最下正周期)
    wcf学习网站
    winform中用户输入查询与拼音首字母的结合,提高用户的操作体验 (转)
    通过SvcUtil.exe生成客户端代码和配置(转)
    WinMail 搭建邮件服务器。
    quick easy ftp server软件在机子上架设了个服务器
  • 原文地址:https://www.cnblogs.com/Bravewtz/p/10624736.html
Copyright © 2020-2023  润新知