nyoj-234-吃土豆（动态规划）

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[n];
	}
	
}