題目234 喫土豆

• 題目234
• 題目信息
• 運行結果
• 本題排行
• 討論區

http://acm.nyist.net/JudgeOnline/problem.php?pid=234

喫土豆

時間限制：1000 ms  |  內存限制：65535 KB

難度：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 ?

輸入

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.

輸出

For each case, you just output the MAX qualities you can eat and then get.

樣例輸入

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

樣例輸出

242

運行號	用戶	題目	結果	時間	內存	語言	提交時間
541212	強子唯一	喫土豆	Accepted	748	8221	java	08-13 16:19:59

import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		while(cin.hasNext()){
			int m =cin.nextInt();
			int n =cin.nextInt();
			int [][]a = new int[m][n];
			for(int i=0;i<m;i++){
				for(int j=0;j<n;j++){
					a[i][j]=cin.nextInt();
				}
			}
			for(int i=0;i<m;i++){
				a[i][1]=Math.max(a[i][0], a[i][1]);
				for(int j=1;j<n-1;j++){
					if((a[i][j+1]+a[i][j-1])>a[i][j]) 
						a[i][j+1]+=a[i][j-1];
					else a[i][j+1]=a[i][j];
				}
			}
			a[1][n-1]=Math.max(a[1][n-1], a[0][n-1]);
			for(int i=1;i<m-1;i++){
				if((a[i-1][n-1]+a[i+1][n-1])>a[i][n-1]) 
					a[i+1][n-1]+=a[i-1][n-1];
				else a[i+1][n-1]=a[i][n-1];
			}
			System.out.println(a[m-1][n-1]);
		}
	}

}