HDU 1087 Super Jumping! Jumping! Jumping!（DP）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1087
题目大意：求最大上升子序列

[align=left]Problem Description[/align]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.
 
[align=left]Input[/align]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.
 
[align=left]Output[/align]For each case, print the maximum according to rules, and one line one case.
 
[align=left]Sample Input[/align]

3 1 3 2
4 1 2 3 4
4 3 3 2 1
0 
[align=left]Sample Output[/align]

4
10
3

思路：基础dp，从前往后转移，if（a[i]>a[j]） dp[i]=max(dp[i],dp[j]+a[i]);  此为状态转移方程
代码：#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
std::ios::sync_with_stdio(false);
int n;
int a[1005];
int dp[1005]; //dp[i]代表以a[i]结束的最大值
while(~scanf("%d",&n))
{
if(n==0)
break;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
a[0]=0;
MEM(dp,0);
for(int i=1;i<=n;i++)
{
for(int j=0;j<i;j++)
{
if(a[i]>a[j])
dp[i]=max(dp[i],dp[j]+a[i]);
}
}
int ans=0;
for(int i=1;i<=n;i++)
ans=max(ans,dp[i]);
printf("%d\n",ans);
}
return 0;
}