Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3011 Accepted Submission(s): 1450

Problem 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

Sample Output

Source 2009 Multi-University Training Contest 4 - Host by HDU
Recommend

先每行求一下最大值，dp1[i] = max(dp[i - 1], dp[i - 2] + cow[i]);
再对这些最大值求一个最大值
dp2[i] = max(dp2[i - 1], dp2[i - 2] + col[i]);

#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; }

hdu2845――Beans

Beans