Super Jumping! Jumping! Jumping!(dp)
Super Jumping! Jumping! Jumping!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32044 Accepted Submission(s): 14425
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.4 1 2 3 4
4 3 3 2 1
0
10
3
题解:求单调上升的序列和的最大值;简单动态规划:
代码:
import java.util.Scanner;
public class hdu1087 {
public static void main(String[] argvs){
Scanner cin = new Scanner(System.in);
while(true){
int n = cin.nextInt();
if(n == )break;
int a[] = new int[n];
int dp[] = new int[n];
int ans = -0x3f3f3f3f;
for(int i = ; i < n; i++){
a[i] = cin.nextInt();
dp[i] = a[i];
for(int j = ; j < i; j++){
if(a[j] < a[i]){
dp[i] = Math.max(dp[j] + a[i], dp[i]);
}
}
ans = Math.max(ans, dp[i]);
}
System.out.println(ans);
}
}
}
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