HDU - 1087 Super Jumping! Jumping! Jumping! （基礎dp）

                           Super Jumping! Jumping! Jumping!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 59387    Accepted Submission(s): 27640

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.

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 2

4 1 2 3 4

4 3 3 2 1

Sample Output

4

10

3

題意：最大遞增子序列之和。

思路：很簡單，貪心一下就好了，dp[ i ] 代表以第 i 位元素結尾時的最大值，枚舉一下 ( 1 ---> i - 1 )，隻要 a[ i ] > a[ j ]， 即 i 元素大于 j 元素，dp[ i ] = max( dp[ i ] , dp[ j ] + a[ i ] ) 即可。 

//#include <bits/stdc++.h>
#include <map>
//#include <tr1/unordered_map>
#include<limits>
#include <float.h>
#include <list>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int lowbit(int x){ return x & (-x); }
inline int read(){int X = 0, w = 0; char ch = 0;while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();return w ? -X : X;}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
#define For(i, x, y) for(int i=x;i<=y;i++)
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x)
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define db double
#define LL long long
#define ULL unsigned long long
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int maxp = 1e3 + 10;
const LL INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int mod = 20090717;

int n, ans;
int a[1010], dp[1010];

int main()
{
    while (~scanf("%d", &n) && n){
        Mem(dp, 0);
        For (i, 1, n) dp[i] = a[i] = read(); //讀入資料，加dp[]數組初始化
        ans = a[1];
        For (i, 2, n)
        For (j, 1, i - 1)
            if (a[i] > a[j])
                dp[i] = max(dp[i], dp[j] + a[i]), ans = max(ans, dp[i]);
        Pri(ans);
    }

    return 0;
}