Beans (DP, two-time DP)

Beans

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) total submission (s): 4141 Accepted Submission (s): 1964

Problem Descriptionbean-eating is a interesting game, everyone owns an m*n matrix, which are filled with different qualiti Es beans. Meantime, there is the only one bean in any 1*1 grid. Now your want to eat the beans and collect the qualities, but everyone must obey by the following rules:if you eat the BEA n at the coordinate (x, y), you can ' t eat the beans anyway at the coordinates listed (if exiting): (x, Y-1), (x, y+1), and The both rows whose Abscissas are x-1 and x+1. Now, how much qualities can do you eat and then get?

Inputthere is a few cases. In each case, there is the integer M (row number) and N (column number). The next M lines each contain N integers, representing the qualities of the beans. We can make sure that the quality of beans isn ' t beyond, and 1<=m*n<=200000.

Outputfor the just output the MAX qualities you can eat and then get.

Sample INPUT4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

Sample Output242

Solving: one-to-row DP, one-to-column DP;

Java timed out ... C would not

Code:

import Java.util.Scanner; Public classBeans {Static intA[] =New int[200010], dp[] =New int[200010], dp[] =New int[200010], b[] =New int[200010];  Public Static voidMain (string[] Argvs) {intM, N; Scanner Cin=NewScanner (System.inch);  while(Cin.hasnext ()) {M=Cin.nextint (); N=Cin.nextint ();  for(inti =0; i < M; i++) {Dp[i]=0;  for(intj =0; J < N; J + +) {Dp[j]=0; A[J]=Cin.nextint (); if(J <2) Dp[j]=A[j]; ElseDp[j]= Math.max (Dp[j-2] + a[j], dp[j-1]); } Dp[i]= Dp[n-1]; if(I <2) B[i]=Dp[i]; ElseB[i]= Math.max (B[i-2] + dp[i], b[i-1]); } System. out. println (B[m-1]); }            }}

Beans (DP, two-time DP)