zoukankan      html  css  js  c++  java
  • [kuangbin带你飞]专题十二 基础DP1 E

    E - Super Jumping! Jumping! Jumping!HDU - 1087

    题目链接:https://vjudge.net/contest/68966#problem/E

    题目:

    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
    题意:给你n个数,求不连续单调递增最长子序列他的和
    思路:dp[i]存i之前的最长单调子序列的和

    //
    // Created by hanyu on 2019/8/4.
    //
    #include <algorithm>
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <set>
    #include<math.h>
    #include<map>
    using namespace std;
    typedef long long ll;
    const int maxn=1e5+7;
    int main()
    {
    
        int n;
        while(~scanf("%d",&n)&&n)
        {
            int a[maxn],dp[maxn];
            memset(dp,0,sizeof(dp));
            memset(a,0,sizeof(a));
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
            }
            for(int i=0;i<n;i++)
            {
                dp[i]=a[i];
                for(int j=0;j<i;j++)
                {
                    if(a[i]>a[j])
                        dp[i]=max(dp[i],dp[j]+a[i]);
                }
            }
            int result=0;
            for(int i=0;i<n;i++)
            {
                result=max(dp[i],result);
            }
            printf("%d
    ",result);
        }
        return 0;
    }
    
    
  • 相关阅读:
    0-Android系统各层中LOG的使用
    Android系统进程Zygote启动过程的源代码分析
    Android应用程序的Activity启动过程简要介绍和学习计划
    分享一个监测企业微信群人员变化的脚本...
    C++ 之 stl::string 写时拷贝导致的问题
    分享一个批量修改文件编码的python脚本
    分享stl sort函数坑点导致coredump问题
    关于使用repo时repo init和repo sync失败的一个解决方案
    sourceinsight sublimetext主题色配置
    父子进程之间的数据拷贝关系
  • 原文地址:https://www.cnblogs.com/Vampire6/p/11299605.html
Copyright © 2011-2022 走看看