HDU 2845 Beans

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2873    Accepted Submission(s): 1392

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

 

Source

2009 Multi-University Training Contest 4 - Host by HDU

解题思路：列的状态只与最多之前三列的选择有关系，滚动数组搞一下求出单的最优值，行的状态与列状态相同，也用滚动数组搞一下，更新答案即可

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int **matrix;int main(){    int m,n,i,j,r[4],c[4],temp,ans;    while(scanf("%d%d",&m,&n)!=-1)    {        ans=-1;        matrix=new int*[m+1];        for(i=0;i<=m;i++)            matrix[i]=new int[n+1];        memset(r,0,sizeof(r));        memset(c,0,sizeof(c));        for(i=1;i<=m;i++)            for(j=1;j<=n;j++)                scanf("%d",&matrix[i][j]);        for(i=1;i<=m;i++)        {            c[1]=matrix[i][1];            c[2]=max(matrix[i][1],matrix[i][2]);            c[3]=max(matrix[i][1]+matrix[i][3],matrix[i][2]);            temp=max(max(c[1],c[2]),c[3]);            for(j=4;j<=n;j++)            {                c[j%4]=max(max(c[(j-2)%4]+matrix[i][j],c[(j-3)%4]+matrix[i][j]),c[(j-1)%4]);                if(temp<c[j%4])                    temp=c[j%4];            }            if(i==1)                r[i%4]=temp;            else if(i==2)                r[i%4]=max(temp,r[1]);            else if(i==3)                r[i%4]=max(r[1]+temp,r[2]);            else                r[i%4]=max(max(r[(i-2)%4]+temp,r[(i-3)%4]+temp),r[(i-1)%4]);        }        for(i=0;i<4;i++)            if(ans<r[i])                ans=r[i];        printf("%d\n",ans);    }    delete matrix;    return 0;}