hdu 1087(LIS变形)

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31458    Accepted Submission(s): 14128

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

 

题意：走一条路 有1-n 个点 ，每个点都有权值,每次走都要比上一次走的点权值大，问最大能够得到多少利益？

题解：LIS变形，求递增的子序列中权值最大的一个，和上次那个LCS变形差不多。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const int N = ;
int dp[N]; ///dp[i]表示到第i点能获得的最大利益
int value[N];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF,n)
    {
        for(int i=; i<=n; i++)
        {
            scanf("%d",&value[i]);
        }
        int mx = dp[] = value[];
        for(int i=; i<=n; i++)
        {
            dp[i] = value[i];
            for(int j=; j<i; j++)
            {
                if(value[j]<value[i]&&dp[i]<dp[j]+value[i])
                {
                    dp[i] = dp[j]+value[i];
                }
            }
            if(mx<dp[i]) mx = dp[i];
        }
        printf("%d\n",mx);
    }
    return ;
}