首页 > ACM题库 > HDU-杭电 > HDU 2845-Beans-动态规划-[解题报告]HOJ
2014
02-17

HDU 2845-Beans-动态规划-[解题报告]HOJ

Beans

问题描述 :

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 ?

输入:

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.

输出:

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.

样例输入:

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

样例输出:

242

 

题意:容易理解。

分析:以后碰到这种类型的题,就要考虑把矩阵先按行来处理,再按列处理。先算出每行能够能够得到的最大值,然后按列处理即可。

代码实现:

#include<stdio.h>
#include<string.h>
int n,m;
int a[200005],dp[200005][2],b[200005];
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int i,j,res;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        res=-1000000;
        dp[0][0]=0;dp[0][1]=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%d",&a[j]);
                dp[j][0]=max(dp[j-1][0],dp[j-1][1]);
                dp[j][1]=dp[j-1][0]+a[j];   
            }
            b[i]=max(dp[m][0],dp[m][1]);
            if(res<b[i])
                    res=b[i];
        }
        dp[0][0]=0;dp[0][1]=0;
        for(i=1;i<=n;i++)
        {
            dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
            dp[i][1]=dp[i-1][0]+b[i];
            if(res<dp[i][0])
                res=dp[i][0];
            if(res<dp[i][1])
                res=dp[i][1];
        }
        printf("%d\n",res);
    }
    return 0;
}

 

解题参考:http://www.cnblogs.com/jiangjing/archive/2013/07/28/3221462.html


  1. 给你一组数据吧:29 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1000。此时的数据量还是很小的,耗时却不短。这种方法确实可以,当然或许还有其他的优化方案,但是优化只能针对某些数据,不太可能在所有情况下都能在可接受的时间内求解出答案。