【 HDU 1087】 Super Jumping! Jumping! Jumping! 动态规划

最新推荐文章于 2024-08-02 10:00:00 发布

原创最新推荐文章于 2024-08-02 10:00:00 发布 · 237 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#HDU 1087

HDU 同时被 2 个专栏收录

18 篇文章

订阅专栏

记忆化搜索

8 篇文章

订阅专栏

本文探讨了一种经典的编程竞赛问题——寻找给定序列中上升子序列的最大和。通过动态规划的方法，文章详细解释了如何求解最长上升子序列之和，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53239 Accepted Submission(s): 24682

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

Sample Output

4 
10 
3

题意：就是给你一个n,然后给你n个数，让你选择一个上升子序列的最大和。

题解:这道题的模型还是LIS问题，只不过我们存储状态dp[i]表示以i结尾的不下降子序列的最大和，预处理的时候，每一个以自身结尾的不下降子序列的最大和就是自己。

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000+7;
int a[maxn],dp[maxn];
int n,ans;
int max(int x,int y)
{
    return x>y?x:y;
}
int LIS()
{
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            if(a[i]<a[j]){
                dp[j]=max(dp[j],dp[i]+a[j]);
            }
        }
    }
    int mx=-1;
    for(int i=1;i<=n;i++)
    mx=max(mx,dp[i]);
    return mx;
}
int main()
{
    while(scanf("%d",&n)==1&&n){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            dp[i]=a[i];
        }
        printf("%d\n",LIS());
    }
    return 0;
}