dp（不连续和）

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 ?

InputThere 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.OutputFor 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
思路：求最大不连续和

#include <iostream>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <set>
using namespace std;
int  dpm[2009][2009] , dpn[2009] ;//dpm求每行的不连续序列和，dpn求每一列的不连续序列和
int len1 , len2 ;

int main()
{

    int n , m ;
    while(cin >> n >> m)
    {
        memset(dpn , 0 , sizeof(dpn));
        memset(dpm , 0 , sizeof(dpm));
        for(int i = 3 ; i < n + 3 ; i++)
        {
            for(int j = 3 ; j < m + 3 ; j++)
            {
                scanf("%d" , &dpm[i][j]);
                dpm[i][j] += max(dpm[i][j-2] , dpm[i][j-3]);//求dp[m+1] 和 dp[m+2]代表两种选择方式最大行不连续和
            }
        }
        for(int i = 3 ; i < n + 3 ; i++)
        {
            dpn[i] = max(dpn[i-2] , dpn[i-3]) + max(dpm[i][m+1] , dpm[i][m+2]);//求不连续最大列序列和
        }
        int ans = 0 ;
        for(int i = 3 ; i < n + 3 ; i++)
        {
            if(dpn[i] > ans)
                ans = dpn[i];
        }
        printf("%d\n" , ans);
    }



    return 0;
}

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