题目

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

翻译

背景

现在,在航电有一种叫做“超级跳！跳！跳！”的游戏非常流行
或许你这么爱学习的孩子对这个游戏知道的不多,我来给你介绍下~

游戏需要不少于两个玩家,包括一个棋盘和一些棋子,并且每一个棋子都有一个正数或者 "start" 、 "end" 标记
玩家从 "start" 标记的棋子开始,结束于 "end" 标记的棋子
玩家可以跳到比当前棋子大的棋子上(将开始棋子看作最小,终点棋子看作最大)
玩家不能往回跳并且可以跳任意的距离
得分为跳到的棋子的总和
也就是说玩家可以直接从起点跳到终点(0分)
你的就是计算出最多能得到多少分(跳多少次)

输入

输入包括多组数据(以 0 结束)
每组数据第一个数 N ( N<=1000 )代表棋子个数
而后 N 个数分别代表棋子上的数(32位的 int )

输出

对于每组数据,在一行内输出最大值

题解

从一组数据中选择出一组递增的子序列,使子序列和最大
也即找出>最大上升子序列<
用 dp[i] 表示跳到第 i 个棋子能达到的最大分数
有 dp[i] = max{ dp[j]+a[i] } (j<i && a[j]<a[i])

最后可以在任一棋子跳出,就需要找出 dp[] 中的最大值

使用最大上升子序列来计算 dp[]

int Max = 0;
a[0] = 0;
for(int i = 0;i <= n;i++) {
    dp[i] = 0;
    for(int j = 0;j < i;j++) 
        if(a[i] > a[j])
            dp[i] = max(dp[i],dp[j] + a[i]);
    Max = max(Max,dp[i]);
}

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
using namespace std;

const int maxn = 1005;
int a[maxn];
int dp[maxn];

bool  Do() {
    int n;
    scanf("%d",&n);
    if(n == 0)
        return false;
    for(int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    int Max = 0;
    a[0] = 0;
    for(int i = 0;i <= n;i++) {
        dp[i] = 0;
        for(int j = 0;j < i;j++) 
            if(a[i] > a[j])
                dp[i] = max(dp[i],dp[j] + a[i]);
        Max = max(Max,dp[i]);
    }
    printf("%d\n",Max);
    return true;
}

int main() {
    while(Do());
    return 0;
}