（最大上升子序列） Super Jumping! Jumping! Jumping! -- hdu -- 1087

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

 

Super Jumping! Jumping! Jumping!

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1087

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. 

 

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. 

 

Output

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

 

Sample Input

3 1 3 2

4 1 2 3 4

4 3 3 2 1

0

 

Sample Output

4

10

3

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
using namespace std;

#define N 1000
#define oo 0x3f3f3f3f

int main()
{
    int n, a[N], sum[N];

    while(scanf("%d", &n), n)
    {
        int i, j;

        memset(a, , sizeof(a));
        memset(sum, , sizeof(sum));

        for(i=; i<=n; i++)
            scanf("%d", &a[i]);

/**

我决定了要好好学 dp 这次写自然是要自己感悟的，自己想真的不容易，
尽管写过，尽管很简单，我也只是无奈，不好想

我又开了个数组 sum[]， 里面记录的是从 1 开始到 i 最大的上升序列的和
既然 sum[i] 里记录的是最大的从 1 到 i 的值，那么每次比较 a[i] 和 a[j] 的大小
如果 a[i] 比 a[j] 大的话，就需要选 sum[i] 和 sum[j]+a[i] 的最大值

最后在 sum 中选个最大值就 OK 啦！！！ 

**/

        for(i=; i<=n; i++)
        {
            sum[i] = a[i];
            for(j=; j<=i; j++)
            {
                if(a[i]>a[j])
                    sum[i] = max(sum[i], a[i]+sum[j]);
            }
        }

        int Max = -oo;

        for(i=; i<=n; i++)
            Max = max(Max, sum[i]);

        printf("%d\n", Max);
    }
    return ;
}