Hdu2845 -- Beans

Hdu2845 -- Beans
BeansTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 3011 Accepted Submission (s): 1450


Problem DescriptionBean-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?
InputThere 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.
OutputFor each case, you just output the MAX qualities you can eat and then get.
Sample Input

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

Sample Output

242

Source2009 Multi-University Training Contest 4-Host by HDU
Recommend

Calculate the maximum value for each line, dp1 [I] = max (dp [I-1], dp [I-2] + cow [I]);
Then, calculate a maximum value for the maximum values.
Dp2 [I] = max (dp2 [I-1], dp2 [I-2] + col [I]);

#include #include 
  
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            #include using namespace std;int cow[200010];int dp[200010];int col[200010];int dp2[200010];int main(){int n, m;while (~scanf("%d%d", &n, &m)){memset (dp2, 0, sizeof(dp2));for (int i = 1; i <= n; ++i){memset (dp, 0, sizeof(dp));for (int j = 1; j <= m; ++j){scanf("%d", &cow[j]);}dp[1] = cow[1];for (int j = 2; j <= m; ++j){dp[j] = max(dp[j - 1], dp[j - 2] + cow[j]);}col[i] = dp[m];}dp2[1] = col[1];for (int i = 2; i <= n; ++i){dp2[i] = max(dp2[i - 1], dp2[i - 2] + col[i]);}printf("%d\n", dp2[n]);}return 0;}