hdu 1087 Super Jumping! Jumping! Jumping!

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19979    Accepted Submission(s): 8670

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

 

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动态规划经典题，题意是找出一个递增数列使其和最大（可不连续）

从左向右处理，对第i项判断其前面的每项是否小于当前项，若小于则说明a[i]可以作为这个序列的递增项，这时只需判断把a[i]加入后当前序列和是否增大

最后只需取出dp数组中的最大值即可

代码如下：

#include <map>#include <cmath>#include <vector>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#define esp 1e-9#define MAXN 1010#define ll long long#define INF 0x7FFFFFFF#define SW(a,b) a^=b;b^=a;a^=b;#define rep(i,j,k) for(int i=j; i<k; ++i)#define REP(i,j,k) for(int i=j; i<=k; ++i)using namespace std;int a[MAXN];int dp[MAXN];int main(void){int n;while(scanf("%d", &n), n){rep(i, 0, n){scanf("%d", &a[i]);}dp[0] = a[0];int max = a[0];rep(i, 1, n){dp[i] = a[i];rep(j, 0, i){if(a[i] > a[j]){//表示取第 i 项前面比 a[i] 小的数 if(dp[i] < dp[j]+a[i])dp[i] = dp[j]+a[i];}}if(dp[i] > max)max = dp[i];}cout << max << endl;} return 0;}