Apple Catching【动态规划-dp】

Apple Catching

 POJ - 2385 

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.

题目大意：开始时站在树1下，第一行输入两个整数n,k，下面的n行代表在1到n秒内每次是哪棵数掉落苹果，且人在两棵树之间可以移动，但最多只能移动k次，问最多能接到几个苹果。

解决方法：dp[i][j]表示的是第i秒移动j次后能接到的最大苹果数，dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])。

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e3 + 5;
const ll mod = 1e9+7;
int arr[maxn];
int dp[maxn][maxn];
int main() 
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(0),cin.tie(0);
    int t,m;
    cin>>t>>m;
    rep(i,1,t) {
        cin>>arr[i];
    }
    int ans=-1;
    for(int i=1;i<=t;i++)
    {
        for(int j=0;j<=m;j++)
        {
            if(j==0)
            {
                if(arr[i]==1)
                    dp[i][j]=dp[i-1][j]+1;    //初始化当一次也不移动时时第i秒的dp
            }
            else
            {
                dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]);    
                if(j%2==0&&arr[i]==1)   //当前在第一棵树下且第一棵树掉落苹果
                    dp[i][j]++;
                else if(j%2==1&&arr[i]==2)    //当前在第二棵树下且第二棵树掉落苹果
                    dp[i][j]++;
            }
            ans=max(dp[i][j],ans);
        }
    }
    cout<<ans<<endl;
    return 0;
}