[dp]poj2385 AppleCatching

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Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8855		Accepted: 4311

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

Sample Output

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.

题意：有两棵苹果树，标号分别为1,2。每分钟有其中的一棵树会掉下一个苹果，奶牛一分钟只能在其中一棵树下接到苹果，但她不知道下一分钟会是那棵树掉下苹果，所以她就要在两棵树下来回跑。但她只会在树下跑W次。问你在T分钟内，奶牛bessie最多能接到多少苹果。

仔细看能看出，是一道dp题，需要递推整个过程才能完成最终的求解。

定义dp[i][j]为奶牛在第i分钟，移动j次以后得到的最大苹果数量

给出状态转移方程：dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+count。这里的dp[i][j]代表在第i分钟移动j次最多能接到的苹果数。在第i分钟奶牛到某棵树下有两种状态：1.从另一棵树走过来(dp[i-1][j-1])2.本来就呆在这棵树下(dp[i-1][j])

在这里我设置了三维数组dp[i][j][k]，第三维度来记录下当前奶牛站的位置(k=0或1)，在做状态转移的时候，只要两种状态都表示出来就可以。在表示时，需要把dp[i][j][0]和dp[i][j][1]存储的最大苹果数量都要记录下来，最后取最大值即可。

最后初始化，根据上面的状态转移方程以及题意，不难想出吧，具体见下面的代码。

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;


int dp[1005][35][2];
//第i分钟移动j次得到最大的苹果数量
//第三维：0 表示站在第一树下
//        1 表示站在第二树下
int num[1005];
int main()
{
    int T,W;
    cin>>T>>W;
    int maxsum = 0;
    for(int i = 1; i <= T; i++)
        scanf("%d",&num[i]);
    for(int i = 1; i <= T; i++)
    {
        dp[i][0][0] = dp[i-1][0][0] + (num[i] ==1);
        dp[i][0][1] = dp[i-1][0][1] + (num[i] ==2);
        for(int j = 1; j <= W; j++)
        {
            dp[i][j][0] = max(dp[i-1][j-1][1],dp[i-1][j][0])+(num[i]==1);
            dp[i][j][1] = max(dp[i-1][j-1][0],dp[i-1][j][1])+(num[i]==2);
            maxsum = max(maxsum,max(dp[i][j][0],dp[i][j][1]));
        }
    }
    cout<<maxsum<<endl;
    return 0;
}