題目：

 

Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 896 Accepted Submission(s): 518   Problem DescriptionNowadays, 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（棋盤）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. InputInput 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. OutputFor 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

Authorlcy

 

題目大意：

求最長上升子序列的最大和。

 

題目分析：

簡單DP。求最長上升子序列的最大和。第一層循環用於遍歷每一個數，用索引i表示。然後第二層循環用於從頭遍歷到當前這個數i的前一個數，用索引j表示。如果data[i] > data[j],則說明以索引j結尾的最長上升子序列加上索引i以後依然能構成一個最長上升子序列，但這是依然需要繼續向後遍歷，因為這是求最長上升子序列的最大和(同一長度的最長上升子序列不唯一)。如果到索引j所形成的最長上升子序列的和大於目前為止的最長上升子序列的和temp，那麼就更新temp的值。然後最後通過temp+data[i]便得到了到索引i所能形成的最長上升子序列的最大和。接下來就是更新一下到目前位置所能得到的全局的最長上升子序列的最大和。(上面所說的那個temp只是局部的最長上升子序列)。

 

至於相求最長上升子序列的長度，可以參考：

 

 

代碼如下：

 

/*
 * c1.cpp
 *
 *  Created on: 2015年2月9日
 *      Author: Administrator
 */


#include 
#include 

using namespace std;

const int maxn = 1005;
int sum[maxn];
int data[maxn];

int n;

int LIS_SUM(){
	int i;
	int j;

	int maxSum = -9999;
	memset(sum,0,sizeof(sum));

	for(i = 1 ; i <= n ; ++i){
		int temp = 0;
		for(j = 1 ; j < i ; ++j){
			if(data[j] < data[i]){
				if(temp < sum[j]){
					temp = sum[j];
				}
			}
		}

		sum[i] = data[i] + temp;

		if(maxSum < sum[i]){
			maxSum = sum[i];
		}
	}

	return maxSum;
}



int main(){
	while(scanf(%d,&n)!=EOF,n){
		int i;
		for(i = 1 ; i <= n ; ++i){
			scanf(%d,&data[i]);
		}

		printf(%d
,LIS_SUM());
	}

	return 0;
}