DESCRIPTION
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 ?
INPUT
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<=200000.
OUTPUT
For each case, you just output the MAX qualities you can eat and then get.
SAMPLE
Input
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
Output
242
题意:给定一个矩阵,每一行中不能选择相邻的数字,不能选择相邻的行,求出所选数的和的最大值。
思路:先对每一行求出其最大不连续和。对每一行所求出的最大值,再求一次最大不连续和。
状态转移方程为
dp[i+1] = max (dp[i], dp[i-1]+num) //表示第 i 个位置所能达到的最大值
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