HDU 2845 Beans

最新推荐文章于 2019-03-28 22:43:44 发布

hqu_fritz

最新推荐文章于 2019-03-28 22:43:44 发布

阅读量444

点赞数

分类专栏： DP

本文链接：https://blog.youkuaiyun.com/hqu_fritz/article/details/39498497

版权

DP 专栏收录该内容

40 篇文章

订阅专栏

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 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

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;
}