HDU2845Beans题解动态规划DP

最新推荐文章于 2018-04-18 08:43:05 发布

原创最新推荐文章于 2018-04-18 08:43:05 发布 · 789 阅读

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#bean #output #each #training #integer #matrix

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本文介绍了一个名为Beans的游戏，玩家在一个M*N的矩阵中收集不同品质的豆子，但需遵循特定规则。文章提供了输入输出样例及算法实现思路，包括状态定义与转移方程。

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Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 616 Accepted Submission(s): 315

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

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gaojie

横竖分别求一下不连续的最大子段和;

状态：

d[i]前i列不连续的最大子段和

s[i]前i行不连续的最大子段和

状态转移方程：

d[j]=max(d[k])+a[i][j] 0<=k<j-1

d[j]+=x;x=max{d[j-1]}

ans=max{s[i]}

代码：

#include<cstdio> #include<algorithm> using namespace std; int main() { int n,m; while(scanf("%d%d",&n,&m)==2) { int i,j,t,y,x=0,r=0,s[5005]={0}; for(i=1;i<=n;i++) { int d[5005]={0}; y=t=0; for(j=1;j<=m;j++) { scanf("%d",d+j); d[j]+=y; t=max(t,d[j]); y=max(y,d[j-1]); } s[i]=x+t; r=max(r,s[i]); x=max(x,s[i-1]); } printf("%d/n",r); } }