POJ 2385 Apple Catching （dp）

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本文探讨了Bessie在两棵苹果树下捕捉掉落苹果的问题，通过动态规划算法解决Bessie如何在有限次移动中捕捉到最多的苹果。文章提供了完整的代码实现。

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Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8363		Accepted: 4093

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

首先本题起始位置为1 那么随着步数的确定 i步的位置必然也是确定的那么状态显而易见

dp[i][j] := 前i步j秒所接的最大苹果数本题采用顺推更加容易写一点

0步在树1下 1步在树2下 2步在树1下 3步在树2下可见二者之和为奇数就能接到苹果那么

dp[i][j + 1] = max(dp[i][j + 1], dp[i][j] + ((i + a[j + 1]) & 1));

dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + ((i + 1 + a[j + 1]) & 1));

AC代码如下：

//
//  POJ 2385 Apple Catching
//
//  Created by TaoSama on 2015-03-04
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#define CLR(x,y) memset(x, y, sizeof(x))

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;

int t, w, a[1005], dp[35][1005];

int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	ios_base::sync_with_stdio(0);

	while(cin >> t >> w) {
		for(int i = 1; i <= t; ++i) cin >> a[i];
		memset(dp, 0, sizeof dp);
		for(int i = 0; i <= w; ++i) {
			for(int j = 0; j <= t; ++j) {
				dp[i][j + 1] = max(dp[i][j + 1], dp[i][j] + ((i + a[j + 1]) & 1));
				dp[i + 1][j + 1] = max(dp[i + 1][j + 1],
				                       dp[i][j] + ((i + 1 + a[j + 1]) & 1));
			}
		}
		int ans = -INF;
		for(int i = 0; i <= w; ++i)
			ans = max(ans, dp[i][t]);
		cout << ans << endl;
	}
	return 0;
}