最长上升子序列问题HDU1087Super Jumping! Jumping! Jumping!

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本文介绍了一款流行棋盘游戏“Super Jumping! Jumping! Jumping!”的玩法及如何通过动态规划算法求解玩家从起点到终点所能获得的最大得分。通过分析棋盘上的棋子分布，使用最长递增子序列的思想来实现。

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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

Sample Output

4
10
3

以dp[i]保存以v[i]为末尾的最长子序列长度，i前面的元素判断是否小于v[i],采用的是顺序结构，也就是i小的元素（前面的元素）已经判断完最长子序列长度，然后i逐渐增大，会调用之前的最长子序列长度，以第一个元素开始产生的最大值与ans比较，大的值付给ans
#include <iostream>
#include <algorithm>
#include <cstdio>
#define maxn 1010
int v[maxn];
long long dp[maxn];
int n;

using namespace std;

//dp[i] 以v[i]为末尾的最长序列长度

void solve()
{
    long long ans = 0;
    for(int i = 0; i < n; i++)
    {
        //以i为末尾的最大值
        dp[i] = v[i];
        for(int j = 0; j < i; j++)
        {
            //如果j元素小于i元素，dp[j] + v[i];
            //j为上一个元素
            //j在前面的循环中作为末尾元素已经进行完判断，选出了到达j的最大值
            if(v[j] < v[i])
            {
                dp[i] = max(dp[i], dp[j] + v[i]);
            }
        }
        ans = max(ans, dp[i]);
    }
    printf("%lld\n", ans);
    return;
}
int main()
{
    while(~scanf("%d", &n))
    {
        if(n == 0)
            break;
        for(int i = 0;i < n; i++)
        {
            scanf("%d", &v[i]);
        }
        solve();
    }
    return 0;
}