hdu 1087 Super Jumping! Jumping! Jumping! （简单 LIS）

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19419 Accepted Submission(s): 8388



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

朴素LIS算法：设f(i)表示L中以ai为末元素的最长递增子序列的长度。时间复杂度O(n^2)
递推思想:在求以ai为末元素的最长递增子序列时，找到所有序号在L前面且小于ai的元素aj，即j<i且aj<ai。如果这样的元素存在，那么对所有aj,都有一个以aj为末元素的最长递增子序列的长度f(j)，把其中最大的f(j)选出来，那么f(i)就等于最大的f(j)加上1，即以ai为末元素的最长递增子序列，等于以使f(j)最大的那个aj为末元素的递增子序列最末再加上ai；如果这样的元素不存在，那么ai自身构成一个长度为1的以ai为末元素的递增子序列。

#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#define N 1005
int a
,f
,ans;
int main()
{
int i,j,n,index;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
f[0]=a[0];
for(i=1;i<=n;i++)
{
index=0;              //寻找满足条件的最大值dp[j]（j<i&&a[j]<a[i]）
for(j=0;j<i;j++)
if(a[j]<a[i]&&f[j]>index)
index=f[j];
f[i]=index+a[i];
}
ans=0;              //答案最小为零
for(i=0;i<n;i++)
ans=ans>f[i]?ans:f[i];
printf("%d\n",ans);
}
return 0;
}