HDU 2845（dp）

题目链接：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): 3097    Accepted Submission(s): 1495

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

 

 

题意：按照规则取出的数字和最大。规则：取了坐标为（x,y）的数字就不能再取坐标为（x,y-1）、（x,y+1）、（x-1,*）和（x+1,*）的数字，*表示任意位置。

分析：首先找出每一行每个位置和最佳情况；每个数字可取可不取，即：dpx[i] = max(dpx[i-1], dpx[i-2] + 该行第 i 个数字)。然后再找列的和最佳情况，此时只有一列，即每一行的最大值组成的列，则：dyp[i] = max(dpy[i-1], dpy[i-2] + 第 i 行的最大值)。

 

#include <cstdio>
#include <cmath>
#include <iostream>
using namespace std;

int n,m,s;
int dpx[222222],dpy[222222];

int main ()
{
    int i,j,k;
    while (scanf ("%d%d",&n,&m)==2)
    {
        memset(dpx, 0, sizeof(dpx));
        memset(dpy, 0, sizeof(dpy));//先对两个数组清零
        for (i=2; i<=n+1; i++)
        {
            for (j=2; j<=m+1; j++)
            {
                scanf ("%d",&s);
                dpx[j] = max(dpx[j-1], dpx[j-2] + s);//每个状态的值等于之前的某个状态加上另一个状态
            }
            dpy[i] = max(dpy[i-1], dpy[i-2] + dpx[m+1]);//因为只有加法，dpx[m+1]为每一行的最大值
        }
        printf ("%d
",dpy[n+1]);
    }
    return 0;
}