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

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

    Sample Output

    4
    10
    3
    题意:求最大上升子序列和
    代码:
     1 #include<cstdio>
     2 #include<algorithm>
     3 using namespace std;
     4 int main()
     5 {
     6     int n,a[1010],dp[1010];
     7     while(scanf("%d",&n)!=EOF&&n)
     8     {
     9         for(int i=0;i<n;i++)
    10             scanf("%d",&a[i]); 
    11         dp[0]=a[0];       
    12         for(int i=0;i<n;i++)
    13         {
    14             dp[i]=a[i];
    15             for(int j=0;j<i;j++)
    16             {
    17                 if(a[i]>a[j])
    18                 {
    19                     dp[i]=max(dp[i],dp[j]+a[i]);
    20                 }
    21             }
    22         }
    23         int ans=0;
    24         for(int i=0;i<n;i++)
    25         {
    26             if(dp[i]>ans)
    27             {
    28                 ans=dp[i];
    29             }    
    30         } 
    31         printf("%d
    ",ans);   
    32     }
    33     return 0;
    34 }
  • 相关阅读:
    纯CSS星级评价
    Enterprise Library启用签名后发生 PublicKeyToken错误,HRESULT:0x80131040解决
    SQL Server
    该如何选择国外VPS
    网站的伪静态化
    kernel FIELD_SIZEOF宏 NULL地址不访问不出错
    Activity的四种加载模式
    Git magic 简短git使用命令集
    为什么包含多句代码的宏要用do while包括起来?
    使用lsof来查看FD和进程的关系
  • 原文地址:https://www.cnblogs.com/hss-521/p/7305793.html
Copyright © 2011-2022 走看看