zoukankan      html  css  js  c++  java
  • HDU 1087 Super Jumping! Jumping! Jumping!(DP)

    Super Jumping! Jumping! Jumping!

    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 2 4 1 2 3 4 4 3 3 2 1 0
     

    Sample Output
    4 10 3
     

    题意   求n个数字的和最大的递增子序列

    基础的dp题目  令d[i]表示以第i个数字结尾的和最大的递增子序列  有d[i]=max(d[i],d[j]+a[i]) j为1到a之间的数  且a[i]>a[j]

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int N = 1005;
    int a[N], d[N];
    int main()
    {
        int ans,n;
        while (scanf ("%d", &n), n)
        {
            ans=0;
            for (int i = 1; i <= n; ++i)
            {
                scanf ("%d", &a[i]);
                d[i]=a[i];
                for(int j=1;j<i;++j)
                    if(a[i]>a[j]) d[i]=max(d[i],d[j]+a[i]);
                ans=max(ans,d[i]);
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    


  • 相关阅读:
    Java 缩放图片工具类,创建缩略图、伸缩图片比例
    93、App Links (应用程序链接)实例
    92、App Permissions(权限管理)实例
    MAC系统教程网站
    Android MVP 详解
    Android开发中,那些让你相见恨晚的方法、类或接口
    用Ultra ISO制作启动U盘装系统
    DB2 Magazine 中文版: 访问 iSeries 数据
    Eclipse快捷键大全
    同步灵无线锂电鼠G11-580HX独特“五灵键”
  • 原文地址:https://www.cnblogs.com/acvay/p/3947296.html
Copyright © 2011-2022 走看看