POJ2385-Apple Catching

这是一个关于算法的问题,涉及到动态规划。问题描述了Bessie在一分钟内从两棵树中捕捉尽可能多的苹果,同时在两棵树之间移动的次数有限制。

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Apple Catching
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12346 Accepted: 5982

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


题意:有两棵苹果树,在t分钟内,每分钟会有一棵树掉下一个苹果,一个人一开始在第一棵树下面,可以从一棵树跑向另一棵树w次,问t时间内他最多能拿到多少个苹果

解题思路:dp,dp[i][j]表示过了i的时间,跑了j次最多能拿到多少苹果


#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[1005], dp[1005][40];

int main()
{
    int t, w;
    while(~scanf("%d%d", &t, &w))
    {
        for(int i = 1; i <= t; i++) scanf("%d", &a[i]);
        memset(dp,0,sizeof dp);
        for(int i = 1; i <= t; i++)
        {
            for(int j = 0; j <= w; j++)
            {
                if(j == 0) dp[i][j] = dp[i-1][j] + a[i]%2;
                else
                {
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]);
                    if(j%2+1 == a[i]) dp[i][j]++;
                }
            }
        }
        printf("%d\n",dp[t][w]);
    }
    return 0;
}

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