nyoj-234-吃土豆(动态规划)
2013-08-13 11:33
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吃土豆
时间限制:1000 ms | 内存限制:65535 KB难度:4
描述 Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want
to eat the beans and collect the qualities, but everyone must obey by the following rules: if you eat the bean 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 you eat and then get ?
输入There are a few cases. In each case, there are two 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 bean isn't beyond 1000, and 1<=M,N<=500.
输出For each case, you just output the MAX qualities you can eat and then get.样例输入
4 6 11 0 7 5 13 9 78 4 81 6 22 4 1 40 9 34 16 10 11 22 0 33 39 6
样例输出
242
来源2009 Multi-University Training Contest 4上传者张洁烽
import java.util.Scanner; public class Main { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Scanner input=new Scanner(System.in); while(input.hasNext()){ int m=input.nextInt(); int n=input.nextInt(); int[] a=new int[n+1]; int[] b=new int[m+1]; for(int i=1;i<=m;i++){ for(int j=1;j<=n;j++) a[j]=input.nextInt(); b[i]=F(a,n); } System.out.println(F(b,m)); } } private static int F(int[] a, int n) { // TODO Auto-generated method stub if(n==1) return a[1]; a[2]=Math.max(a[1], a[2]); if(n==2) return a[2]; for(int i=3;i<=n;i++){ a[i]+=a[i-2]; if(a[i]<a[i-1]) a[i]=a[i-1]; } return a ; } }
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