首页 > 代码库 > hdu2845——Beans
hdu2845——Beans
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 ?
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
242
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]);
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 <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
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。