POJ：2385-Apple Catching（dp经典题）

Apple Catching

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 14311 Accepted: 7000

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.

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解题心得：

1. 题意就是有两颗苹果树，每一秒钟某一颗树上会掉下来一颗苹果，一个人可以在两棵树下移动，移动时间不计，最多可以移动ｋ次，问这个人最多可以获得多少颗苹果。
2. 其实这个题最直观的做法就是用记忆化搜索，直接按照题意搜索就行了。当然也可以将记忆化搜索提炼出ｄｐ公式出来，写ｄｐ状态转移方程式。
3. 提炼出来ｄｐ方程式可以这样表示，dp[i][j]为在ｉ秒最多移动ｊ次可以获得的最大苹果树。状态转移方程为dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]) + (j%2+1 == arr[i])，因为一开始在１位置，所以在移动ｊ次之后所在的位置为ｊ％２＋１，代表在第ｉ秒移动ｊ次的状态只能够从第ｉ－１秒移动ｊ－１次和ｉ－１秒移动ｊ次转移而来，然后加上当前是否可以接到苹果的。

记忆化搜索代码：

#include <algorithm>
#include <stdio.h>
#include <cstring>
using namespace std;
const int maxn = 1010;
int dp[maxn][35][3],w,n,scor[maxn];

void init() {
    memset(dp,-1,sizeof(dp));
    scanf("%d%d",&n,&w);
    for(int i=1;i<=n;i++) {
        scanf("%d",&scor[i]);
        scor[i] %= 2;
    }
}

int dfs(int times,int change,int pos) {
    if(times > n || change > w)
        return 0;
    if(dp[times][change][pos] != -1)
        return dp[times][change][pos];
    return dp[times][change][pos] = ((scor[times] == pos) + max(dfs(times+1,change,pos),dfs(times+1,change+1,!pos)));
}

int main() {
    init();
    int ans = max(dfs(1,0,1),dfs(1,0,0));
    printf("%d\n",ans);
    return 0;
}

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dp方程式转移代码

#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 1010;
int dp[maxn][35];
int arr[maxn];

int main() {
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d",&arr[i]);
    for(int i=1;i<=n;i++) {
        for(int j=0;j<=k;j++) {
            if(j == 0) {
                dp[i][j] = dp[i-1][j] + (j%2 +1 == arr[i]);
                continue;
            }
            dp[i][j] = max(dp[i-1][j],dp[i-1][j-1]);
            if(j%2 + 1 == arr[i])
                dp[i][j]++;
        }
    }

    int ans = 0;
    for(int i=0;i<=k;i++)
        ans = max(ans,dp[n][i]);

    printf("%d\n",ans);
    return 0;
}