hdu1087 Super Jumping! Jumping! Jumping!--DP
发布日期:2021-10-03 20:31:51
浏览次数:3
分类:技术文章
本文共 2340 字,大约阅读时间需要 7 分钟。
原题链接:
一:原题内容
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 24 1 2 3 44 3 3 2 10
Sample Output
4103
二:分析理解
一个简单的动态规划,起初理解成一个最大连续递增子序列,结果错了,题意: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.这句话。后来改了下,加个dp数组,其中dp[i]表示以a[i]结尾的最大分数。
三:AC代码
#include#include #include using namespace std;int a[1010];int dp[1010];int main(){ int N; while (scanf("%d", &N) && N) { for (int i = 0; i < N; i++) scanf("%d", &a[i]); int max = a[0]; dp[0] = a[0]; for (int i = 1; i < N; i++)//求出dp[1]到dp[N-1] { dp[i] = a[i]; for (int j = 0; j < i; j++)//求出dp[i] { if (a[j] < a[i]) if (dp[j] + a[i]>dp[i]) dp[i] = dp[j] + a[i]; } if (dp[i] > max) max = dp[i]; } printf("%d\n", max); } return 0;}
转载地址:https://blog.csdn.net/LaoJiu_/article/details/50936369 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!
发表评论
最新留言
哈哈,博客排版真的漂亮呢~
[***.90.31.176]2024年04月17日 03时56分57秒
关于作者
喝酒易醉,品茶养心,人生如梦,品茶悟道,何以解忧?唯有杜康!
-- 愿君每日到此一游!
推荐文章
VTK:复合数据之Generate3DAMRDataSetWithPulse
2019-04-26
VTK:复合数据之MultiBlockDataSet
2019-04-26
VTK:复合数据之OverlappingAMR
2019-04-26
VTK:数据结构之BuildLocatorFromKClosestPoints
2019-04-26
VTK:数据结构之BuildOctree
2019-04-26
VTK:数据结构之ClosestNPoints
2019-04-26
IDEA SpringMVC Tomcat 404 一网打尽 及路径问题汇总【踩坑记录】
2019-04-26
boost::bad_function_call用法的测试程序
2019-04-26
SpringMVC框架学习总结
2019-04-26
boost::function_types::is_function用法的测试程序
2019-04-26
boost::geometry::clear用法的测试程序
2019-04-26
asp.net mvc 视图赋值有哪几种
2019-04-26
继承的使用要点
2019-04-26
asp.net mvc 传参和视图赋值--model传参
2019-04-26
从客户端中检测到有潜在危险的Request.Form值
2019-04-26