Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40650 Accepted Submission(s): 18805
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
题解：
该题的意思是求一个递增最大和序列
给出n，在给出n个数，然后从左往右跳，规则是必须跳向比自己大的数，只能往右跳，中间可以隔着数跳
问这些数加起来，最大和是多少
下面是自己举的例子和演算过程
arr[]:存放n个数 dp[][]:存放和
在dp[i][i]中存放在跳到arr[i]这个数时，之前的最大和是多少
判断arr[i]这一步的最大和时，先查找之前比arr[i]小的数，若存在arr[j]
即dp[i][j] = dp[j][j] + arr[i];
然后再将dp[i][]这一行找到的最大值放在dp[i][i]处，以便后续使用
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 5 using namespace std; 6 7 typedef long long LL; 8 LL dp[1010][1010]; 9 10 int main() 11 { 12 int n; 13 LL arr[1010]; 14 while(scanf("%d",&n) && n) { 15 16 for(int i = 0 ; i < n ; i++) 17 scanf("%lld",&arr[i]); 18 LL ans; 19 for(int i = 0; i < n ; i++){ 20 ans = arr[i]; 21 for(int j = 0; j < i; j++) { 22 if(arr[i] > arr[j]){ 23 dp[i][j] = dp[j][j] + arr[i]; 24 ans = ans < dp[i][j] ? dp[i][j] : ans; 25 } 26 else 27 dp[i][j] = arr[i]; 28 } 29 dp[i][i] = ans; 30 } 31 for(int i = 0 ; i < n ;i++) 32 ans = ans < dp[i][i] ? dp[i][i] : ans; 33 printf("%lld ",ans); 34 } 35 return 0; 36 }