Super jumping! jumping! jumping!Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total Submission (s): 25057 Accepted Submission (s): 11073
Problem descriptionnowadays, a kind of chess game called "Super jumping! jumping! Jumping! "is very popular in HDU. Maybe you is a good boy, and know little about the this game, so I introduce it to you now.
The game can be played by and more than the players. It consists of a chessboard (chessboard) and some chessmen (chess pieces), and all chessmen is marked by a positive integer or "start" or "End ”. The player starts from start-point and must jumps to end-point finally. In the course of jumping, the player would visit the chessmen in the path, but everyone must jumps from one Chessman to Ano Ther absolutely bigger (you can assume start-point are 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 can straightly get to end-point From Start-point. Of course you get the 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.
Inputinput contains multiple test cases. Each test case was described in a line as follow:
N value_1 value_2 ... value_n
It is guarantied, that N was not more than, and all value_i be in the range of 32-int.
A test case, starting with 0 terminates, the input and this test are not processed.
Outputfor, print the maximum according to rules, and one line one case.
Sample Input
3 1 3 24 1 2 3 44 3 3 2 10
Sample Output
4103
Title Link: vhttp://acm.hdu.edu.cn/showproblem.php?pid=1087
The main topic: find the length of the sequence of n maximum increment subsequence and (number does not require continuous), such as example one: Increment subsequence has (1,) (1,3) (3) (2), and the largest is (1,3), and for 4.
title resolution: Dp,dp[i] represents the maximum increment subsequence of a sequence consisting of a first I number containing a[i]. Initialize Dp[i]=a[i], find the number a[j] smaller than the previous A[i], the state transfer return equation is:Dp[i]=max (Dp[i],dp[j]+a[i]).
#include <cstdio> #include <cstring>const int maxn=1005;int A[MAXN],DP[MAXN]; Dp[i] represents the largest sub-sequence and int main () {int N,i,j,max, which contains the first I position of the first I number ; while (scanf ("%d", &n)!=eof&&n) { for (i=0;i<n;i++) scanf ("%d", &a[i]); MAX=DP[0]=A[0]; for (i=1;i<n;i++) { dp[i]=a[i]; Initialize Dp[i]for (j=i-1;j>=0;j--) { if (A[j]<a[i]&&a[i]+dp[j]>dp[i]) //state transition equation: Dp[i] =max (A[i]+dp[j],dp[i]) { dp[i]=dp[j]+a[i]; if (Dp[i]>max) max=dp[i];} } } printf ("%d\n", max); } return 0;}
The code is as follows:
HDU 1087 Super jumping! jumping! jumping! (DP)