Super Jumping! Jumping! Jumping! (经典dp最大升序字段和)
来源:互联网 发布:美联储非农就业数据 编辑:程序博客网 时间:2024/05/22 12:36
Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1268 Accepted Submission(s): 690Problem 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
//动态规划问题就是通过前一状态来解决现在状态。//当前点的值,通过与前面点的状态来求解现在的状态最大和。//#include <iostream>#include <algorithm>#include <string.h>#include <cmath>#include <stdio.h>#include <vector>#include <map>#include <queue>#include <utility>using namespace std;int a[100005];int dp[100005];int main(){ int n; while (cin >> n && n) { memset(dp, 0, sizeof(dp)); memset(a, 0, sizeof(a)); for (int i = 1; i <= n; ++i) cin >> a[i]; int ans = dp[1] = a[1]; for (int i = 2; i <= n; ++i) { dp[i] = a[i]; for (int j = 1; j < i; ++j) { if (a[i] > a[j]) dp[i] = max(dp[i], dp[j] + a[i]); ans = max(ans, dp[i]); } } cout << ans << endl; }}
0 0
- Super Jumping! Jumping! Jumping! (经典dp最大升序字段和)
- 1087Super Jumping! Jumping! Jumping(寻找和最大的升序子序列)
- hdoj1087 Super Jumping! Jumping! Jumping!(DP)
- HDU1087-Super Jumping! Jumping! Jumping!(dp)
- hdu Super Jumping! Jumping! Jumping!(DP)
- hdu 1087 Super Jumping! Jumping! Jumping!(最大递增字段和)
- Super Jumping! Jumping! Jumping!(DP)
- 【dp】Super Jumping! Jumping! Jumping!
- Super Jumping! Jumping! Jumping!DP
- Super Jumping! Jumping! Jumping! --DP
- Super Jumping! Jumping! Jumping! dp
- Super Jumping! Jumping! Jumping! dp
- hdu 1087 Super Jumping! Jumping! Jumping!(dp:上升子序列最大和)
- HDOJ/HDU 1087 Super Jumping! Jumping! Jumping!(经典DP~)
- hduoj Super Jumping! Jumping! Jumping!【最大上升子序列和】
- hdu1087 - Super Jumping! Jumping! Jumping! (dp 求递增子序列的最大和)
- hdu 1087 Super Jumping! Jumping! Jumping!(基础DP,最大上升子序列和)
- hdoj 1087 Super Jumping! Jumping! Jumping! 【dp&&最大递增子段和】
- 欢迎使用CSDN-markdown编辑器
- AC自动机专题——J
- How to configure a SMTP server in Red Hat Enterprise Linux
- 动态规划练习-1(最长上升子序列)
- httpclient 两种方式请求网络
- Super Jumping! Jumping! Jumping! (经典dp最大升序字段和)
- mysql之视图
- 数组的连续最大子段和
- 计算机网络总结之计算机概述
- requre.js 用法
- python控制电脑Beep()蜂鸣函数制作简易播放器
- 在中国技术进阶为管理是一条正常的发展道路么?
- flex的row和column居中
- bzoj1195(最短母串)