怒刷DP之 HDU 1087
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
System Crawler (2015-09-05)
Description
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
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
Sample Input
4 1 2 3 4
4 3 3 2 1
0
Sample Output
10
3
#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using namespace std; const int SIZE = ;
const int INF = 0x7ffffff0;
struct Node
{
int val,max;
}DP[SIZE]; int main(void)
{
int n,ans,box,max; while(scanf("%d",&n) != EOF && n)
{
ans = -INF;
for(int i = ;i < n;i ++)
{
scanf("%d",&box);
max = ;
for(int j = ;j < i;j ++)
if(DP[j].max < box && DP[j].val > max)
max = DP[j].val;
DP[i].max = box;
DP[i].val = max + box;
ans = ans > DP[i].val ? ans : DP[i].val;
}
printf("%d\n",ans);
} return ;
}
怒刷DP之 HDU 1087的更多相关文章
- 怒刷DP之 HDU 1257
最少拦截系统 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Statu ...
- 怒刷DP之 HDU 1160
FatMouse's Speed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Su ...
- 怒刷DP之 HDU 1260
Tickets Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Stat ...
- 怒刷DP之 HDU 1176
免费馅饼 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status ...
- 怒刷DP之 HDU 1114
Piggy-Bank Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit S ...
- 怒刷DP之 HDU 1069
Monkey and Banana Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u S ...
- 怒刷DP之 HDU 1024
Max Sum Plus Plus Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u S ...
- 怒刷DP之 HDU 1029
Ignatius and the Princess IV Time Limit:1000MS Memory Limit:32767KB 64bit IO Format:%I64d &a ...
- 【DP】HDU 1087
HDU 1078 Super Jumping! Jumping! Jumping! 题意: 有这么个游戏,从start到end(自己决定在哪停下来)连续跳圈,中间不能空一个圈不跳,圈里的数字必须比你上 ...
随机推荐
- java提高数据库访问效率代码优化
package com.jb.jubmis.comm; import java.sql.Connection;import java.sql.ResultSet;import java.sql.SQL ...
- EXTJS项目实战经验总结一:日期组件的change事件:
1 依据选择的日期,加载相应的列表数据,如图: 开发说明 1 开发思路: 在日期值变化的事件中获得选择后的日期值,传给后台,然后从后台加载相应的数据 2 问题:在查看extjs2.2 的a ...
- X431 元征诊断枪
X-431 Diagun是专门为汽车维修技师设计的诊断设备. 小巧的主机.强大的诊断功能.方便快捷的网上升级.一体化多功能接头,都是维修技师的首选.X-431 Diagun 是汽车维修技师的标准装备. ...
- PHP中关于超链接的拼接问题
<?php$link = " http://www.baidu.com";echo '<a href='.$link.'> 百度 </a>';?> ...
- js页面文字选中后分享到新浪微博实现
demo您可以狠狠地点击这里:js文字选中分享到新浪微博demo 方法与代码 选中即分享的功能看上去比较高级,其实实现是相当简单的.其中的会让人头大,一般人也不感兴趣的原理这里就直接跳过.这个js文字 ...
- ArcSDE 10.2建立SDE服务
从ArcGIS 10.1开始,arcgis官方推荐以直连方式连接SDE,因此在SDE安装时不再自动安装SDE服务,以下是手动安装SDE服务的方法 环境 服务端: oracle 11.2 64位,Arc ...
- hdu 5258 数长方形 离散化
数长方形 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5258 Des ...
- linux C(undefined reference to `sqrt')
那是因为没有链接到math库 可以这样来做,在后面加上-lm. 代码如下: gcc 10.c -o 10 -lm
- 有用好看的CSS+JS+table 导航
预览效果图 watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dis ...
- 离线安装Android开发环境的方法
对于大家从官网上下载下来的SDK其实是一个安装工具,里面啥都没有,如果在线安装的话会需要很长时间.我们同样可以从网络上用下载工具将所需要安装的东西下载下来,(同样有劳大家自己动手找找了)然后直接放入相 ...