dp跳棋
来源:互联网 发布:游戏爆率算法 编辑:程序博客网 时间:2024/04/28 11:21
Problem D
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 34 Accepted Submission(s) : 14
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
思路分析:
逐个计算能到每点的最大和值,最后检索最大值。
代码实现:
#include <stdio.h>#include <string.h>int main() {int n, i, j, max, array[1010], sum[1010];while(scanf("%d",&n)) {if (n == 0) break;memset(array, 0, sizeof(array));//初始化memset(sum, 0, sizeof(sum));for (i = 0;i < n;i ++) scanf("%d", &array[i]);sum[0] = array[0];for (i = 0;i < n;i ++) {if (i == 0) {continue;}max = 0;for (j = 0;j < i;j ++) {//循环找出最大值赋入if (array[i] > array[j] && max < sum[j])max = sum[j];}sum[i] = array[i] + max;}for (max = 0,i = 0;i < n;i ++) if (max < sum[i]) max = sum[i];printf("%d\n", max);}
}
- dp跳棋
- 跳棋
- 跳棋
- 跳棋
- 跳棋
- 跳棋
- 跳棋
- 跳棋jump
- HDU1087 跳棋
- 国际跳棋
- C#编写跳棋程序
- 跳棋被搞定了
- 跳棋棋盘图片
- C/C++跳棋问题
- 西洋跳棋走棋算法
- 国际跳棋的开局
- (搜索)跳棋系列2
- 南邮 OJ 1733 跳棋
- 面向对象编程初步
- Java基础10-接口,抽象类以及异常
- tomcat启动报错 java.net.SocketException: select failed
- TCO2012Round1A-2-EllysFractions
- 今天所遇到的问题
- dp跳棋
- 与未知同行:论敏捷开发中的反馈与反复
- memcache for python2.6
- 模式分解——适配器模式初探
- 有点害怕了,知耻后勇
- 一种简单的密码验证方案
- 杭电ACM题目分类
- Java基础11-包,jar以及Eclipse
- STL概述《using stl》