题目234 吃土豆

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• 题目234
• 题目信息
• 运行结果
• 本题排行
• 讨论区

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

     
运行号
用户
题目
结果
时间
内存
语言
提交时间
541212
强子唯一
吃土豆
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]);}}}

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