hdu 1087 Super Jumping! Jumping! Jumping! ( 求序列的最大上升序列 )
来源:互联网 发布:数据字典是系统中各类 编辑:程序博客网 时间:2024/05/28 11:51
Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17746 Accepted Submission(s): 7617
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.
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.
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
Author
lcy
题意:求序列的最大上升序列,(可跳)不用连续。
分析:开另一个数组b来保存同下标的“最大值”,每一个都要向前找合法的最大b值。
感想:最近在刷dp啊dp....从水题开始感悟啊感悟~~~~~~~~
代码:
#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;int a[1001],b[1001];int main(){ int n,i,j; while(scanf("%d",&n)&&n) { memset(b,0,sizeof(b)); for(i=1;i<=n;i++) scanf("%d",&a[i]); for(i=1;i<=n;i++) { int maxb=0; b[i]=a[i]; for(j=1;j<=n;j++) { if(a[j]<a[i]) { maxb=max(maxb,b[j]); //printf("test:%d %d \n",j,maxb); } } b[i]+=maxb; } int sum=0; for(i=1;i<=n;i++) sum=max(sum,b[i]); printf("%d\n",sum); } return 0;}
- hdu 1087 Super Jumping! Jumping! Jumping! ( 求序列的最大上升序列 )
- hdu 1087 Super Jumping! Jumping! Jumping!(LIS变形题,求上升子序列的最大和)
- HDU 1087 Super Jumping!Jumping!Jumping求连续上升子序列的最大和值 (解析)
- HDU 1087 Super Jumping! Jumping! Jumping!(求最大上升子序列和)
- HDU 1087 Super Jumping! Jumping! Jumping!最长上升子序列
- HDU 1087 Super Jumping! Jumping! Jumping!(求绝对递增子序列的最大和)
- (hdu step 3.2.3)Super Jumping! Jumping! Jumping!(DP:求最长上升子序列的最大和)
- hdu 1087 Super Jumping! Jumping! Jumping!(上升子序列最大和)
- hdu 1087 Super Jumping! Jumping! Jumping! 最大上升子序列。模板题
- hdu 1087 Super Jumping! Jumping! Jumping!(基础DP,最大上升子序列和)
- HDU 1087 Super Jumping! Jumping! Jumping! (线性dp 最大上升子序列)
- hdu 1087 Super Jumping! Jumping! Jumping!(dp:上升子序列最大和)
- Hdu 1087 Super Jumping! Jumping! Jumping!【最大上升子序列和】
- hdu 1087 Super Jumping! Jumping! Jumping!(最大上升子序列之和)
- HDU 1087 Super Jumping! Jumping! Jumping! 上升序列最大和+DP .
- HDU 1087 Super Jumping! Jumping! Jumping!(最大上升子序列)
- HDU-1087 Super Jumping! Jumping! Jumping!(上升子序列最大和)
- HDU 1087 Super Jumping! Jumping! Jumping!(最大上升子序列和)
- 安装rpm包出现信赖错误(/bin/sh)
- 驱动层HOME按键的配置
- js的值传递和引用传递
- android:onTouch()和onTouchEvent()的区别?看完这篇文章就知道了
- 如何调试SIGABRT和EXC_BAD_ACCESS引起的crash
- hdu 1087 Super Jumping! Jumping! Jumping! ( 求序列的最大上升序列 )
- 关于shared pool的深入探讨
- MyEclipse中快捷键的使用
- SPring Bean注入 job
- 【Lucene3.6.2入门系列】第09节_高级搜索之自定义QueryParser
- TranslateAnimation.setAnimationListener 内部方法不执行
- Android从相机或相册获取图片裁剪
- SQL 獲取一年的每一日.每週,某個星期的數據
- 对 Linux 新手有用的 20 个命令