Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31490    Accepted Submission(s): 14146

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
DP水题。。。
 #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = ;
int a[maxn];
long long dp[maxn];
void solve(){
int n;
while(scanf("%d",&n)!=EOF&&n){
for(int i = ; i<=n; i++) scanf("%d",&a[i]);
memset(dp,,sizeof(dp));
long long ans = ;
for(int i = ; i<=n; i++){
dp[i] = a[i];
for(int j = ; j<=i; j++){
if(a[i]>a[j]){
dp[i] = max(dp[i],dp[j]+a[i]);
}
}
ans = max(ans,dp[i]);
}
printf("%I64d\n",ans);
}
}
int main()
{
solve();
return ;
}

Super Jumping! Jumping! Jumping! 基础DP的更多相关文章

  1. 「kuangbin带你飞」专题十二 基础DP

    layout: post title: 「kuangbin带你飞」专题十二 基础DP author: "luowentaoaa" catalog: true tags: mathj ...

  2. 基础dp

    队友的建议,让我去学一学kuangbin的基础dp,在这里小小的整理总结一下吧. 首先我感觉自己还远远不够称为一个dp选手,一是这些题目还远不够,二是定义状态的经验不足.不过这些题目让我在一定程度上加 ...

  3. 基础DP(初级版)

    本文主要内容为基础DP,内容来源为<算法导论>,总结不易,转载请注明出处. 后续会更新出kuanbin关于基础DP的题目...... 动态规划: 动态规划用于子问题重叠的情况,即不同的子问 ...

  4. hdu 5586 Sum 基础dp

    Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Desc ...

  5. hdu 4055 Number String (基础dp)

    Number String Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)To ...

  6. 训练指南 UVA - 10917(最短路Dijkstra + 基础DP)

    layout: post title: 训练指南 UVA - 10917(最短路Dijkstra + 基础DP) author: "luowentaoaa" catalog: tr ...

  7. 训练指南 UVA - 11324(双连通分量 + 缩点+ 基础DP)

    layout: post title: 训练指南 UVA - 11324(双连通分量 + 缩点+ 基础DP) author: "luowentaoaa" catalog: true ...

  8. M - 基础DP

    M - 基础DP Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Descriptio ...

  9. lightoj1004【基础DP】

    从低端到顶端求个最大值: 思路: 基础DP,递推 #include<cstdio> #include<queue> #include<map> #include&l ...

随机推荐

  1. Github建站全攻略

    本系列为原创,发表在我的github主页,详细介绍了如何在github上建立个人网站,还包括域名绑定.评论功能.站内搜索等辅助功能的介绍,欢迎交流.   一步步在GitHub上创建博客主页(6) 本篇 ...

  2. process lasso 优化原理

    <星际争霸2:虚空之遗>很多玩家的CPU性能并不低,但是在星际2中的表现就总会出现掉帧的情况,那么应该如何提升CPU的性能就成了玩家关注的话题,下面小编就为大家带来星际争霸2虚空之遗cpu ...

  3. 使用onclick跳转到其他页面。使用button跳转到指定url

    1. onclick="javascript:window.location.href='aa.htm'" 2.  onclick="location='URL'&quo ...

  4. QQ登录界面

    @property (nonatomic,assign) IBOutlet UITextField *qq; @property (nonatomic,assign) IBOutlet UITextF ...

  5. 贪心+bfs 或者 并查集 Codeforces Round #268 (Div. 2) D

    http://codeforces.com/contest/469/problem/D 题目大意: 给你一个长度为n数组,给你两个集合A.B,再给你两个数字a和b.A集合中的每一个数字x都也能在a集合 ...

  6. 查看SQL Server 2008的版本及位数

    如何查看SQL Server 2008的版本及位数及SP版本: 登录SQL Server,找到“SQL查询分析器”,输入“Select @@version”,运行,即可看出版本及SP版本. 该方法适用 ...

  7. MySql 加锁问题

    1.设置非自动提交 set autocommit=0;  这时候 for update才会起作用 2.一般用法 set autocommit=0;  for update(加锁)  ;  commit ...

  8. sql查询百分号的方法

    select * from [tablename] where [col] like '%100/%%' escape '/'

  9. 【转】使用ThinkPHP必须掌握的调试方法

    经常看到有人问到findAll的返回数据类型是什么之类的问题,以及出错了不知道什么原因的情况,其实还是没有熟悉ThinkPHP内置的调试手段和方法,抛开IDE本身自带的调试方式不说,如果你正在用或者打 ...

  10. 最小点集覆盖/HDU2119

    题目连接 先试一下题/?/ 最小点集覆盖=最大匹配 /*根据i.j建图,跑一边最大匹配 */ #include<cstdio> #include<cstring> using ...