HDU2845--Beans

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

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

Sample Output

242

 

/*首先求出每行我们能取得最大值*/#include <iostream>#include <cstdio>using namespace std;#define maxn 200008int key[maxn];int sum[maxn];//用来存每一行能取得的最大值int dp[maxn];inline int max(int a,int b){    return a>b?a:b;}int main(){    int r,c;    while(scanf("%d%d",&r,&c)==2)    {        for(int i=1;i<=r;i++)        {            for(int j=1;j<=c;j++)            {                scanf("%d",&key[j]);                if(j==1)dp[j]=max(0,key[j]);                else                {                    dp[j]=max(dp[j-1],dp[j-2]+key[j]);                }            }            sum[i]=dp[c];        }        //好吧，这样我就求出每行能拿的最大值了        dp[1]=sum[1];        for(int i=2;i<=r;i++)        {            dp[i]=max(dp[i-1],dp[i-2]+sum[i]);        }        printf("%d\n",dp[r]);    }    return 0;}