POJ 2385 Apple Catching(DP)

Apple Catching Time Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

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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.

用opt[i][j]表示第i分钟走了j次后最多能摘几个苹果

由于只有两棵树，故用j%2+1就能判断此时Bessie在哪棵树下面，省下了一维空间

f[i]表示第i分钟哪颗树掉苹果，若f[i]==j%2+1,则此时可以摘得苹果，opt[i][j]=max(opt[i-1][j],opt[i-1][j-1])+1

否则，此时摘不到苹果，opt[i][j]==max(opt[i-1][j],opt[i-1][j-1])

但对于opt[i][0]要单独考虑，即第i分钟时仍然没移动，在第一棵树下的情况，可由opt[i-1][0]推导

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int dp[1001][31],f[1001];
int main()
{
    int t,w;
    while(~scanf("%d %d",&t,&w))
    {
        for(int i=1;i<=t;i++) scanf("%d",&f[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=t;i++){
            dp[i][0]=dp[i-1][0];
            if(f[i]==1)
                dp[i][0]++;
            for(int j=1;j<=w&&j<=i;j++){
                if(j%2+1==f[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]);
            }
        }
        int k=dp[t][0];
        for(int i=1;i<=w;i++)
           k=max(k,dp[t][i]);
        printf("%d\n",k);
    }
    return 0;
}