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

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087

    题意分析:LIS变形,由于是要求最大和上升子序列而不是最长上升子序列,故状态转移方程:dp[i] = max(dp[i], dp[j]+a[i])

    /*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
     
    
    Author
    lcy
    */
    
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    using namespace std;
    const int maxn = 1000 + 10;
    int a[maxn], n, dp[maxn];
    int LIS()
    {
        int ans = 0;
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i < n; i++){
            dp[i] = a[i];
            for(int j = i-1; j >= 0; j--){
                if(a[i] > a[j])
                    dp[i] = max(dp[i], dp[j] + a[i]);
            }
            ans = max(ans, dp[i]);
        }
        return ans;
    }
    int main()
    {
        while(~scanf("%d", &n) && n != 0){
            for(int i = 0;i < n; i++) scanf("%d", a+i);
            int cnt = LIS();
            printf("%d
    ", cnt);
        }
        return 0;
    }
  • 相关阅读:
    大神语录1 如何滑动fragmentmanager里面一个fragment里面的viewgroup---dispatchTouchEvent 、onInterceptTouchEvent 、onTouchEvent
    转载-好的简历
    安卓开发8- 安卓开源项目整理github
    安卓开发7-网络通信-如何使用webservice
    安卓开发6 -Viewpager+fragment更新数据
    leetcode rotatenumber
    Java程序执行时间的简单方法
    LeetCode happyint
    安卓开发5-fragment和activity
    [转]深入理解AsyncTask的工作原理
  • 原文地址:https://www.cnblogs.com/ACFLOOD/p/4296442.html
Copyright © 2011-2022 走看看