hdu 2845 Beans dp

题意：n*m的格子取数，每次取x，y时，x+1和x-1这两行不能去，（x，y-1），（x，y+1）这两格不能取。求最后能取到的最大的值。

先对每行求最大不连续子段和，由于行数不能连续取，相等于再对列做一次求不连续子段和即可

#include <bits/stdc++.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
typedef long long ll;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 200010;

int a
, b
, dp
;
int n, m;

int main()
{
	while( ~scanf("%d%d", &n, &m) ) {
		for( int i = 1; i <= n; ++i ) {
			for( int j = 1; j <= m; ++j ) {
				dp[j] = 0;
				scanf("%d", &a[j]);
			}
			dp[0] = 0, dp[1] = a[1];
			for( int j = 2; j <= m; ++j ) {
				dp[j] = max( dp[j-2]+a[j], dp[j-1] );
			}
			b[i] = dp[m];
		}
		memset( dp, 0, sizeof(dp) );
		dp[1] = b[1];
		for( int i = 2; i <= n; ++i ) {
			dp[i] = max( dp[i-2]+b[i], dp[i-1] );
		}
		printf("%d\n", dp
);
	}
	return 0;
}