HDU 1087 Super Jumping! Jumping! Jumping!

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.

 

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

 

动态规划： 

这道题要求的是最长上升子序列的和。用一个 s 数组存储子问题的解。（s对应的下标 i 表示，从0到i，最长上升子序列的和。）从头开始，遍历每一个数，开始第一重循环。对每一个数，又从头遍历到这个数的前一个数，这是第二重循环，在第二重循环中，用下标j表示当前遍历到的数，如果 a[i] > a[j]，则说明 a[j] 和 a[i] 能构成一个上升子序列，但此时还需继续往后遍历（因为要求最长），用tem保存最大的s[j]（到第j个数时的解，最大的s[j]才能保证最长子序列）。遍历完后，将此时的s[i]更新（s[i] += a[i]+term），解决当前的子问题。

 

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define maxn 1010

 4 

 5 int a[maxn], s[maxn];

 6 int main()

 7 {

 8     int n;

 9     while(scanf("%d", &n) != EOF){

10         int i, j, maxx = 0;

11         if(n == 0)

12             break;

13         for(i = 0; i < n; i++)

14             scanf("%d", &a[i]);

15         memset(s, 0, sizeof(s));

16 

17         for(i = 0; i < n; i++){

18             int temp = 0;

19             for(j = 0; j < i; j++){

20                 if(a[j] < a[i]){

21                     if(temp < s[j])

22                         temp = s[j];

23                 }

24             }

25             s[i] += a[i] + temp;

26             if(maxx < s[i])

27                 maxx = s[i];

28         }

29         printf("%d\n", maxx);

30     }

31     return 0;

32 }