POJ 2385 Apple Catching

最新推荐文章于 2020-08-19 09:18:00 发布

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最新推荐文章于 2020-08-19 09:18:00 发布

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原文链接：http://www.cnblogs.com/cjoier-nfy/p/7436695.html

在一个有趣的场景中，奶牛Bessie试图捕捉从两棵树上掉落的苹果。受到行动限制，Bessie必须在有限时间内高效地移动并捕捉尽可能多的苹果。通过动态规划算法，本文详细介绍了如何计算Bessie能够捕捉的最大苹果数。

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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.

设计状态dp[i][j]代表到I时间转换了j次的最大收益，那么状态显然可以由之前的状态转还是不转转移过来，注意加上一段区间的值可以用前缀和来优化。

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<stack>
 9 #include<queue>
10 #include<map>
11 #define RG register
12 #define IL inline
13 #define pi acos(-1.0)
14 #define ll long long 
15 using namespace std;
16 int tree1[1005],tree2[1005];
17 int dp[1005][35];
18 int T,W;
19 int main() {
20   scanf("%d%d",&T,&W);
21   for(int i=1;i<=T;i++){
22     int type;
23     scanf("%d",&type);
24     if(type==1) tree1[i]=1;
25     else tree2[i]=1;
26     tree1[i]+=tree1[i-1];
27     tree2[i]+=tree2[i-1];
28   }
29   for(int i=1;i<=T;i++)
30     for(int j=0;j<i;j++){
31       for(int w=0;w<=W;w++){
32         if(w%2==0){
33        dp[i][w]=max(dp[i][w],dp[j][w]+tree1[i]-tree1[j]);
34        if(w+1<=W) dp[i][w+1]=max(dp[i][w+1],dp[j][w]+tree2[i]-tree2[j]);
35     } else{
36        dp[i][w]=max(dp[i][w],dp[j][w]+tree2[i]-tree2[j]);
37        if(w+1<=W) dp[i][w+1]=max(dp[i][w+1],dp[j][w]+tree1[i]-tree1[j]);
38     }
39       }
40     }
41   int maxn=0;
42   for(int i=0;i<=W;i++) maxn=max(maxn,dp[T][i]);
43   cout<<maxn;
44   return 0;
45 }