POJ 2385 Apple Catching

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Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15231		Accepted: 7465

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 22112211

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  

題意有兩棵樹1，2，掉蘋果，人一開始在1下，有指定的移動次數，給蘋果下落，求最多接到的蘋果數

首先想一想，跟平常的題有什麼區別

1.人會移動，

2.蘋果要判斷是否要不移動接到，還是移動接到

3.移動的次數不像是背包問題裡的背包容量越多越好，但和背包問題類似

for(int i=1;i<=n;i++)    {        dp[i][0]=dp[i-1][0];//第i個蘋果，j次移動的最好結果，判斷一直不動的情況        if(t[i]==1) dp[i][0]++;                for(int j=1;j<=w;j++)        {            if(j%2+1==t[i]){//判斷移動後的位置在哪，如果跟這次下落蘋果的位置一樣                dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+1;//dp的關鍵，跟上次的東西有關聯，接這個蘋果我可以選擇i-1個蘋果移動相同的次數，然後不動接
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　//也可以從另一個蘋果樹移動過來接            }            else            dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);        }    }

整體代碼

#include<iostream>#include<string.h>#include<algorithm>using namespace std;int t[1000];int dp[1000][1000];int main(){    int n,w;    cin>>n>>w;    for(int i=1;i<=n;i++)    {        cin>>t[i];    }        memset(dp,0,sizeof(dp));    for(int i=1;i<=n;i++)    {        dp[i][0]=dp[i-1][0];        if(t[i]==1) dp[i][0]++;                for(int j=1;j<=w;j++)        {            if(j%2+1==t[i]){                dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+1;            }            else            dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);        }    }    cout<<dp[n][w]<<endl; } 

POJ 2385 Apple Catching