Apple Catching

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本文探讨了一道经典的算法问题——Bessie如何在两棵树间移动以抓住最多数量的苹果，同时限制移动次数。通过动态规划的方法给出了问题的解决方案，并提供了详细的代码实现。

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Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14191		Accepted: 6938

Description

It is a little known fact that cows love apples. Farmer John has two apple trees (which are conveniently numbered 1 and 2) in his field, each full of apples. Bessie cannot reach the apples when they are on the tree, so she must wait for them to fall. However, she must catch them in the air since the apples bruise when they hit the ground (and no one wants to eat bruised apples). Bessie is a quick eater, so an apple she does catch is eaten in just a few seconds.

Each minute, one of the two apple trees drops an apple. Bessie, having much practice, can catch an apple if she is standing under a tree from which one falls. While Bessie can walk between the two trees quickly (in much less than a minute), she can stand under only one tree at any time. Moreover, cows do not get a lot of exercise, so she is not willing to walk back and forth between the trees endlessly (and thus misses some apples).

Apples fall (one each minute) for T (1 <= T <= 1,000) minutes. Bessie is willing to walk back and forth at most W (1 <= W <= 30) times. Given which tree will drop an apple each minute, determine the maximum number of apples which Bessie can catch. Bessie starts at tree 1.

Input

* Line 1: Two space separated integers: T and W

* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.

Output

* Line 1: The maximum number of apples Bessie can catch without walking more than W times.

Sample Input

Sample Output

Hint

INPUT DETAILS:

Seven apples fall - one from tree 2, then two in a row from tree 1, then two in a row from tree 2, then two in a row from tree 1. Bessie is willing to walk from one tree to the other twice.

OUTPUT DETAILS:

Bessie can catch six apples by staying under tree 1 until the first two have dropped, then moving to tree 2 for the next two, then returning back to tree 1 for the final two.

Source

USACO 2004 November

思路：

定义dp[i][j]：在第i棵树时，步数为j时取的最大苹果数。

状态转移时分为两种情况：

1、从第i棵树，步数为j时转移过来。苹果数+1；

2、从另一棵树棵树，步数为j-1转移过来。苹果数+1；

两种情况取最优即可。

特判：

由于题目规定起点在第一棵树，那么取第一个苹果时就需要特判一下：如果第一个苹果在地一棵树，那么dp[0][0]=1；如果在第二棵树，那么dp[1][1]=1；有了这个初始状态，后面就可以放心的转移了。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define inf 1<<29
#define mod 1000000000
int n,m,a[1005],dp[3][1005],ans;
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof dp);
        int ans=0;
        
        for(int i=0;i<n;i++)
        {
            if(i==0)
            {
                if(a[0]==1) dp[0][0]=1;
                else dp[1][1]=1;
                continue;
            }
            for(int j=m;j>=0;j--)
            {
                if(j)
                    dp[a[i]-1][j]=max(dp[a[i]-1][j],dp[a[i]%2][j-1])+1;
                else
                    dp[a[i]-1][j]=dp[a[i]-1][j]+1;
                ans=max(ans,dp[a[i]-1][j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}