POJ 2385 Apple Catching 接苹果 DP

浏览数：36 / 时间：2015年06月11日

Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7858		Accepted: 3846

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.

Source

USACO 2004 November

题意：

两棵树，在一定的时刻会掉下苹果，有一头牛初始站在第一棵树下，它能在两棵树下来回移动，但步数有限制。问它最多能接到多少个苹果。

分析：

在某一时刻牛的状态为，它可能上一时刻它也在这棵树下，或者是从另外一棵树移过来的。这样，我们用dp[i][j]来表示在第i个时刻走了j步接到的最多苹果数。暂且不管当前这一时刻是否接到苹果，dp[i][j] = max(dp[i-1][j], dp[i-1][j-1])，这描述了我前面说的两种状态。

再看当前这一秒是否接到苹果。因为初始在第一棵树下，因而如果j为奇数，即移动了奇数步，那么它应该在第一棵树下，此时如果第一棵树下会掉下一个苹果，那么dp[i][j]就要加1。同理如果移动偶数步且第i时刻第二棵树上掉下一个苹果，dp[i][j]也要加1。

代码：

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

const int MAX_T = 1010;
const int MAX_W = 33;
int T, W, a[MAX_T];
int dp[MAX_T][MAX_W];
void solve() {
    for(int i = 0; i <= W; i++) dp[0][i] = 0;
    for(int i = 1; i <= T; i++) {
        dp[i][0] = dp[i-1][0]+2-a[i];
        for(int j = 1; j <= W && j <= i; j++) {
            dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]);
            if(j % 2) dp[i][j] += a[i]-1;
            else dp[i][j] += 2-a[i];
        }
    }
    int cnt = 0;
    for(int i = 0; i <= W; i++)
        if(dp[T][i] > cnt) cnt = dp[T][i];
    printf("%d\n", cnt);
}
int main() {
    while(~scanf("%d%d", &T, &W)) {
        for(int i = 1; i <= T; i++)
            scanf("%d", &a[i]);
        solve();
    }
    return 0;
}

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POJ 2385 Apple Catching 接苹果 DP