hdu 2845 Beans

题目：

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

 

参 考 ：http://972169909-qq-com.iteye.com/blog/1447073

题 意：

　　在图中取数，例如取了81之后，同一行的相邻两个不能取，还有81的上面那行和下面那行也不能取，问能取到的最大和是多少？

做法 ：

　　最大连续子串和

代码：

　　

//最大连续子串和

#include<iostream>
#include<stdio.h>
using namespace std;

const int T=200005;

int main( ){

    int n,m,A[T],B[T];
    while(~scanf("%d %d",&n,&m)){
        for( int i=1;i<=n;i++){
            for( int j=1;j<=m;j++){
                scanf("%d",&A[j]);
                if( j==2 ) A[j]=max( A[j], A[j-1] );
                else if( j>2 )  A[j]=max( A[j]+A[j-2], A[j-1] );
            }
            if( i==1 )   B[i]=A[m];
            else if(i==2) B[i]=max( A[m],B[i-1]);
            else B[i]=max ( A[m]+B[i-2], B[i-1]);
        }
        cout << B[n] <<endl;
    }
    return 0;
}