首页 > ACM题库 > HDU-杭电 > HDU 1087 Super Jumping! Jumping! Jumping!-动态规划-[解题报告] C++
2013
11-27

HDU 1087 Super Jumping! Jumping! Jumping!-动态规划-[解题报告] C++

Super Jumping! Jumping! Jumping!

问题描述 :

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.

输入:

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.

输出:

For each case, print the maximum according to rules, and one line one case.

样例输入:

3 1 3 2
4 1 2 3 4
4 3 3 2 1
0

样例输出:

4
10
3

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1087

题意:只能走比当前旗子大的旗子,不能回头,求走过最大的旗子的和。

mark:简单dp。

代码:

#include <stdio.h>
#include <string.h>

int main()
{
    int n,a[1010],dp[1010][2];
    int i,j,max;
    while(scanf("%d", &n), n)
    {
        memset(dp, 0, sizeof(dp));
        for(i = 0; i < n; i++)
            scanf("%d", a+i);
        for(i = max = 0; i < n; i++)
        {
            dp[i][0] = dp[i][1] = a[i];
            for(j = 0; j < i; j++)
                if(dp[j][0] < dp[i][0])
                    if(dp[i][1] < dp[j][1]+a[i]) dp[i][1] = dp[j][1]+a[i];
            if(max < dp[i][1]) max = dp[i][1];
        }
        printf("%d\n", max);
    }
    return 0;
}

  1. 第一句可以忽略不计了吧。从第二句开始分析,说明这个花色下的所有牌都会在其它里面出现,那么还剩下♠️和♦️。第三句,可以排除2和7,因为在两种花色里有。现在是第四句,因为♠️还剩下多个,只有是♦️B才能知道答案。

  2. 有两个重复的话结果是正确的,但解法不够严谨,后面重复的覆盖掉前面的,由于题目数据限制也比较严,所以能提交通过。已更新算法

  3. Thanks for using the time to examine this, I truly feel strongly about it and enjoy finding out far more on this subject matter. If achievable, as you achieve knowledge

  4. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。