题意：如图中所示，选一个数后，它左右两边的数 和 上下2行的数不能再选，问这样选出的数和的最大值思路：对每一行求出：这一行选数字的最大值（也就是 i行 做一个最大值dp[i] ，则这一行的最大值是 dp[i][m] )对列来求dp时 ，已知每一行的最值是 dp[i][m]，则这里做一个小dp2mark#include #include #define N 2000in
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2021-08-13 13:40:28
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#include#define N 200100int f[N];int a[N],n;int main() {int m,j,i,suma,sumb,sumc,sumd;while(scanf("%d%d",&m,&n)!=EOF) { for(i=1;isumb) sumc=suma+f[(...
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2013-11-06 18:28:00
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本来是很简单的一道题，却想了好长时间由于数据量比较大，所以逐行读入，逐行处理先处理每一行的不相邻元素和的最大值，记录在数组b中最后计算不相邻行的和的最大值二者的状态转移方程都类似：dp[j] = max(dp[j - 1], dp[j - 2] + a[j]); 1 //#define LOCAL ...
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2014-07-27 09:22:00
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链接:http://acm.hdu.edu.cn/showproblem.php?pid=2845 Beans Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 973  
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2011-08-17 10:26:42
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思路：动态规划 对于一行来说，相邻的数不可同时取，容易得到状态转移方程： dp[i]=max(dp[i-2]+a[i],dp[i-1]); 然后取每一行的最大得b[i] ,又可看作一行相同处理
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2023-03-03 12:42:43
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BeansTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1433Accepted Submission(s): 741Problem 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
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2021-07-29 16:24:58
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BeansTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1819Accepted Submission(s): 921 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 be
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2013-03-24 22:56:00
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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2845分析：题目意思就是不能有任小，即c[i]=c[i]+max(c[i-2],c[i-3])；那么对于每一列也同样考虑.
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2013-05-10 21:24:00
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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 intere
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2023-04-24 08:36:42
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Description
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2023-02-03 10:02:07
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Problem Description Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there i
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2016-03-25 16:19:00
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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 t
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2015-07-17 19:51:00
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小记：之前因为N*M<= 20W，所以我是使用
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2023-07-18 18:40:28
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思路：二维动态规划，先求每一行的最大和，组成新的一
#include
int r[200010][2],c[200010][2];
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
main()
{
int i,j,R,C,s;
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2023-06-09 18:31:57
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BeansTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2596Accepted Submission(s): 1279Problem Descr...
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2015-10-11 17:45:00
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本文是一篇关于java独立的帖子BeansTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1945
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2013-05-24 21:46:00
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原题链接： http://acm.hdu.edu.cn/showproblem.php?
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2022-12-06 19:16:17
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dp[i] = max ( dp[i-2] + num[i] , dp[i-1] )没什么好解释的，经典模板题#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define MAX 200007
using namespace
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2023-04-24 01:47:15
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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2845思路：对于一行来说，相邻的数不可同时取，容易得到状态转移方程：sum[i] = max (sum[i-2]+sum[i], sum[i-1])，其中sum[i]表示一行前i个数时的最大和；然后把sum[m]保存到另一个数组中，对于每一行都这么做，然后最后在对数组再次进行一边这样的操作就行了；View Code 1 #include<iostream> 2 const int N=200020; 3 using namespace std; 4 5 int a[N],b[N]; 6
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2013-03-12 21:42:00
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题目地址：点击打开链接
思路：参考了别人的思想，开了四个数组A了，想想其实开3个数组就行了，结果看大神的代码开了2个直接A了，没看懂。
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2022-08-04 14:09:37
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