POJ2385——Apple Catching

                                            $Apple~Catching$

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16248		Accepted: 8009

$Description$

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.

$Sample~Input$

7 2
2
1
1
2
2
1
1

$Sample~Output$

6

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

 

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解法一：

　　用一个数组$f_{i,j}$表示第$i$时刻，走了$j$步的最多苹果数。

　　状态转移方程为：

　　　　从第一棵苹果树出发：

　　　　　　$$f_{i,j}=max\{f_{i-1,j},f_{i-1,j-1}\}+(j~\mod~2)= a_i$$

　　　　从第二棵苹果树出发：

　　　　　　$$f_{i,j}=max\{f_{i-1,j},f_{i-1,j-1}\}+(j~\mod~2)\neq a_i$$

　　$code:$

#include <cstdio>
#include <cstring>
#define max(a,b) ((a)>(b)?(a):(b))
using namespace std;

int read()
{
	int x=0,f=1;char c=getchar();
	while (c<'0' || c>'9'){if (c=='-')f=-1;c=getchar();}
	while (c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-48;c=getchar();}
	return x*f;
}

const int MAXN=1005;
const int MAXM=45;
int n,m;
int a[MAXN];
int f[MAXN][MAXM];

int main()
{
	n=read();m=read();
	for (int i=1;i<=n;i++) a[i]=read()-1;
	memset(f,0,sizeof(f));
	for (int i=1;i<=n;i++)
		for (int j=0;j<=m;j++)
			f[i][j]=max(f[i-1][j],f[i-1][j-1])+((j&1)==a[i]);
	int ans=0;
	for (int i=0;i<=m;i++)
		ans=max(ans,f[n][i]);
	memset(f,0,sizeof(f));
	for (int i=1;i<=n;i++)
		for (int j=0;j<=m;j++)
			f[i][j]=max(f[i-1][j],f[i-1][j-1])+((j&1)!=a[i]);
	for (int i=0;i<=m;i++)
		ans=max(ans,f[n][i]);
	printf("%d\n",ans);
	return 0;
}

解法二：

　　用一个数组$f_{i,j}$表示移动了$i$步，当前位置在第$j$棵苹果树下的时候的最多苹果数。

　　对于第$k$棵树上掉下的一个苹果，要么是之前就已经移动$i$步到了第$k$棵树并等到苹果掉下，或者是移动$i-1$步，到另一棵树下，现在赶到这棵树下。状态转移方程为：

　　　　$$f_{i,j}=max\{f_{i,j},f_{i-1,(j+1)\mod~2}\}+1$$

　　$code:$

#include <iostream>
#include <cstdio>
using namespace std;

int read()
{
  	int x=0,f=1;char c=getchar();
  	while (c<'0' || c>'9'){if (c=='-')f=-1;c=getchar();}
  	while (c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-48;c=getchar();}
  	return x*f;
}
int n,w,come;
int a[31][2]={0};

int main()
{
    n=read();w=read();
    for(int i=1;i<=n;i++)
    {
        come=read()-1;
        for (int j=0;j<=w;j++)
        	a[j][come]=max(a[j][come]+1,a[j-1][(come+1)%2]+1);
    }
    printf("%d\n",max(a[w][0],a[w][1]));
    return 0;
}