HDU1087 Super Jumping! Jumping! Jumping!(动态规划，最大递增子段和)

题目：

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35426    Accepted Submission(s): 16112

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 24 1 2 3 44 3 3 2 10

 

Sample Output

4103

 

Author

lcy

 

Recommend

 

Statistic | Submit | Discuss | Note思路：

给了一串数字，让你找出最大递增的字段和，在这个题中dp[i]代表以当前数字为最后一个数字所能达到的最大和，把ans初始化为一个极小值，以当前位置为基准，从0遍历到这个数，每个数都进行比较，每一步更新dp【i】的值，最后在dp中找出最大的值就行。

代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int a[1008];int dp[1008];//dp[i]表示在第i个位置，所能达到的最大和int main(){    int n;    while(scanf("%d",&n)&&n)    {        int ans;        mem(dp,0);        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        for(int i=1; i<=n; i++)        {            ans=-9999;            for(int j=0; j<i; j++)                if(a[i]>a[j])                    ans=max(ans,dp[j]);//更新ans            dp[i]=ans+a[i];        }        ans=-9999;        for(int i=0; i<=n; i++)            if(dp[i]>ans)                ans=dp[i];//找出最大的        printf("%d\n",ans);    }    return 0;}