Beans HDU

题目链接:点我

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 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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

Sample Output

242

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题意:

一个 n * m的矩阵,矩阵的每个点有不同的值,你每次可以选取坐标为(x, y) 的矩阵但是选取了这个点后你不能选取它左右相邻的点,也不能选取上一行和下一行,问最大值是多少.

思路:

动态规划, 对于一行来说，相邻的数不可同时取，容易得到状态转移方程：
dp[i]=max(dp[i-2]+a[i],dp[i-1]); 每一行和每一列进行相同的处理.

代码:

#include<cstdio>#include<algorithm>#include<cmath>#include<iostream>using namespace std;const int maxn = 200000+10;int dpx[maxn],dpy[maxn];int main(){    int n,m;    while(scanf("%d %d", &n, &m) != EOF){        for(int i = 2;i<=n+1;++i){            for(int  j = 2;j<=m+1;++j){                int s;                scanf("%d", &s);            dpy[j]=max(dpy[j-1],dpy[j-2]+s);            }            dpx[i]=max(dpx[i-1],dpx[i-2]+dpy[m+1]);        }        printf("%d\n",dpx[n+1]);    }    return 0;}