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.
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.
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.
3 1 3 2 4 1 2 3 4 4 3 3 2 1 0
4 10 3
【题解】 题意很简单，下棋，有起点start和终点end，每个人都必须从起点出发，每次可以跳到周围的棋子上，可以相邻，也可以不相邻（即一次可以跳很远），但前提是要比当前的棋子值大，得分是每次跳到的棋子上的数值之和，现在求最大数值和，简单分析一下就知道这是求个最长上升子序列的和。
【AC代码】
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1005;
int dp[N];
int m,a[N];int main()
{while(~scanf("%d",&m),m){for(int i=0;i<m;i++)scanf("%d",&a[i]);memset(dp,0,sizeof(dp));for(int i=0;i<m;i++){for(int j=0;j<i;j++){if(a[i]>a[j]){dp[i]=max(dp[i],dp[j]+a[i]);}}dp[i]=max(dp[i],a[i]);//一定要有这个 这是保证如果都是下降序列的话取第一个数值}int ans=0;for(int i=0;i<m;i++)ans=max(ans,dp[i]);printf("%d\n",ans);}return 0;
}