hdu2845 Beans

最新推荐文章于 2019-01-28 00:54:01 发布

原创最新推荐文章于 2019-01-28 00:54:01 发布 · 1.6k 阅读

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文章标签：

#bean #output #each #training #integer #matrix

本文介绍了一款名为“豆子游戏”的问题及其解决方案。该问题是寻找在一个M*N矩阵中，按照特定规则能够获取的最大豆子价值。文章通过动态规划的方法，首先计算每一行的最大非连续选择值，然后再次应用动态规划计算这些最大值的整体最优解。

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

Recommend

gaojie

简单的dp，对每行找一遍不连续取的最大值，之后对这些最大值再找一遍不连续的最大值即可。

笔误，WA一次。

#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;

typedef struct
{
    int choose;
    int notch;
}DP;

vector <int> a[200005];
DP dp[200005];
int sum[200005];

int main()
{
    int i,j,n,m,x,T;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for (i=0;i<n;i++)
        {
            a[i].clear();
            for (j=0;j<m;j++)
            {
                scanf("%d",&x);
                a[i].push_back(x);
            }
        }
        memset(sum,0,sizeof(sum));
        for (i=0;i<n;i++)
        {
            dp[0].choose=a[i][0];
            dp[0].notch=0;
            for (j=1;j<m;j++)
            {
                if (j==1) dp[j].choose=a[i][1];
                else dp[j].choose=max(dp[j-2].choose,dp[j-2].notch)+a[i][j];
                dp[j].notch=max(dp[j-1].choose,dp[j-1].notch);
            }
            sum[i]=max(dp[m-1].choose,dp[m-1].notch);
        }
        dp[0].choose=sum[0];
        dp[0].notch=0;
        for (i=1;i<n;i++)
        {
            if (i==1) dp[i].choose=sum[i];
            else dp[i].choose=max(dp[i-2].choose,dp[i-2].notch)+sum[i];
            dp[i].notch=max(dp[i-1].choose,dp[i-1].notch);
        }
        printf("%d\n",max(dp[n-1].choose,dp[n-1].notch));
    }
    return 0;
}