hdu2845之Beans,DP
2016-07-07 21:30
253 查看
hdu2845之Beans,DP
2012-12-08 16:12 533人阅读 评论(0) 收藏 举报
版权声明:本文为博主原创文章,未经博主允许不得转载。
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
242
先求某一行能获得的最大值,然后求所有行能获得的最大值.
2012-12-08 16:12 533人阅读 评论(0) 收藏 举报
版权声明:本文为博主原创文章,未经博主允许不得转载。
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
242
先求某一行能获得的最大值,然后求所有行能获得的最大值.
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<queue> #include<algorithm> #include<map> #include<iomanip> #define INF 99999999 using namespace std; const int MAX=200003; int dpx[MAX],dpy[MAX];//dpx用来存当前这一行前i列最大值,dpy用来存前i行可得最大值. int main(){ int n,m,s; while(cin>>n>>m){ for(int i=2;i<=n+1;++i){ for(int j=2;j<=m+1;++j){ scanf("%d",&s); dpx[j]=max(dpx[j-1],dpx[j-2]+s);//当前这行前j列能获得的最大值. } dpy[i]=max(dpy[i-1],dpy[i-2]+dpx[m+1]);//前i行最大能获得的最大值. } cout<<dpy[n+1]<<endl; } return 0; }
相关文章推荐
- 39. Combination Sum
- SQL
- Codeforces Round #361 (Div. 2) D RMQ+二分
- 简单的二进制
- java之原子性
- 网站web.cofig配置用户的权限
- hadoop入门级总结一:HDFS
- hadoop入门级总结一:HDFS
- Tomcat配置文件常用操作
- Hive建表、加载数据模板
- 13_03_Linux进程管理之一
- OSI-L3协议ARP
- HDU 3714 Error Curves(模拟退火刷时间)
- PAT L2-004. 这是二叉搜索树吗?
- Java 8 指南
- Python Tutorial 学习笔记1
- 一个颜色器的demo
- 各种排序算法比较
- POJ-1144 Network((割顶)
- OSI-VLSM(Variable Length Subnet Mask)可变长子网掩码