題意：

給你兩個(gè)數(shù)字n和m；代表會(huì)有n個(gè)蘋果掉落，m次可以移動(dòng)的機(jī)會(huì)；有兩棵樹，開始你站在樹1下面，一分鐘只能移動(dòng)一次，下面的數(shù)值代表在哪一顆樹下會(huì)掉落蘋果；問你在可移動(dòng)的范圍內(nèi)，最多可以接到多少個(gè)蘋果？

題目：

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.

分析：

(1）.基礎(chǔ)dp題,按照題意很容易看出有兩個(gè)變量，分鐘和移動(dòng)機(jī)會(huì)，所以我們按照習(xí)慣寫出兩層for循環(huán)，簡(jiǎn)單思考其中的關(guān)系，就可以得到dp定義,$dp[i][j]$,表示在i時(shí)間內(nèi),用 $j$次轉(zhuǎn)移機(jī)會(huì)得到的最大蘋果數(shù).
(2).因?yàn)殚_始的位置在第一棵樹下，那么我們就可以由移動(dòng)的步數(shù) $j$的奇偶性判斷現(xiàn)在在哪顆樹下；
(3).dp轉(zhuǎn)移如下,如果 $j==0$,則 $dp[i][j]=dp[i?1][j]$,考慮j為0時(shí)，到第i分鐘，一步都沒有走動(dòng)。
否則 $dp[i][j]=max(dp[i?1][j],dp[i?1][j?1])$.就是當(dāng)前從上一個(gè)狀態(tài)下，走還是不走獲得蘋果最多。
(4)如果現(xiàn)在所在的樹的編號(hào)和現(xiàn)在掉落的蘋果所在的位置相同，那么接到的蘋果的數(shù)量姐增加，由（2）這個(gè)我就放在最后判斷一下就行，也可以放在第三步，直接在推導(dǎo)式里面。
(5).最后在 $dp[n][i]$里找最大值就行了,$(0<=i<=n).$表示一共走i步時(shí)，n分鐘吃到蘋果最大值。
我今天真的閑的羅里吧嗦這么多，自戀的覺得只要跟著我的思路走，一定能會(huì)這道題，hhh

AC模板：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int M=1e3+10; int n,m,ans; int s[M],dp[M][35]; int main() {while(~scanf("%d%d",&n,&m)){ans=0;memset(dp,0,sizeof(dp));for(int i=1; i<=n; i++)scanf("%d",&s[i]);if(s[1]==1)dp[1][0]=1;elsedp[1][1]=1;for(int i=2; i<=n; i++)for(int j=0; j<=i&&j<=m; j++){if(j==0)//考慮j為0時(shí)，也就是說，到第i分鐘，一步都沒有走動(dòng)。dp[i][j]=dp[i-1][j]+s[i]%2;else{dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);//就是當(dāng)前走還是不走獲得蘋果最多。if(j%2+1==s[i])//因?yàn)樽铋_始站在樹1下面，可以用走多少步的奇偶來表示當(dāng)前在那棵樹下。dp[i][j]++;}}for(int i=0; i<=m; i++)//表示一共走i步時(shí)，n分鐘吃到蘋果最大值。ans=max(ans,dp[n][i]);printf("%d\n",ans);}return 0; }

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