標籤:dp
Beans
Time 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
先每行求一下最大值,dp1[i] = max(dp[i - 1], dp[i - 2] + cow[i]);
再對這些最大值求一個最大值
dp2[i] = max(dp2[i - 1], dp2[i - 2] + col[i]);
#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>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