HDU 1087 Super Jumping! Jumping! Jumping! 简单DP
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1087
题目大意：N个数字组成的一条路，每个数字是一个节点顺序连接，起点在第一个数之前，终点在第N个数之后。现让你选择一条路从起点跳到终点，只能往前且跳到比当前点大的那个点，可以认为终点是最大的，可以从起点直接跳到终点但是路的值就是0了，每条路的值为所经过的数字节点的值的和，问你值最大为多少。
解题思路：决策：在当前点i往i~N哪个点跳，反过来当前点i+1可以从1~i哪个点跳过来，那么a[i+1] > a[i]，很明显：
dp[i]：=走到第i个点的最大值
dp[i] = max(dp[j] + a[i]) 1 < j < i, a[i] > a[j]
代码：
1 const int maxn = 1e3 + 5; 2 int n; 3 ll a[maxn]; 4 ll dp[maxn]; 5 6 void solve(){ 7 memset(dp, 0, sizeof(dp)); 8 for(int i = 1; i <= n; i++) dp[i] = a[i]; 9 ll ans = 0; 10 for(int i = 2; i <= n; i++){ 11 for(int j = 1; j < i; j++){ 12 if(a[j] < a[i] && dp[j] + a[i] > dp[i]){ 13 dp[i] = dp[j] + a[i]; 14 } 15 } 16 ans = max(ans, dp[i]); 17 } 18 printf("%I64d\n", ans); 19 return; 20 } 21 int main(){ 22 while(scanf("%d", &n) && n){ 23 for(int i = 1; i <= n; i++) scanf("%I6d", &a[i]); 24 solve(); 25 } 26 }
题目：
Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40679 Accepted Submission(s): 18824
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.
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.
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
0
Sample Output
4
10
3
