Apple Catching

Apple Catching

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 4729
Accepted: 2261

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 22112211

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.

Source

USACO 2004 November



解法：
#include <iostream>
#include <cstdio>
using namespace std;

int max(int a,int b)
{
  if(a>b)
    return a;
  else return b;
}

int main()
{

  int n,m;
  int dp[1001][31][2];
  int a[1001];
  scanf("%d%d",&n,&m);


  //初始化工作：
  for(int i=1;i<=n;i++)
    {
  scanf("%d",&a[i]);
    }
  if(a[1]==1)
    {
 dp[1][0][1]=1;
 dp[1][1][1]=1;//就算第一步转移到树2的话，仍然为1个
    }
  else
    {
      dp[1][0][2]=1;
      dp[1][1][2]=1;
    }


  //核心：
  for(int i=2;i<=n;i++)
    {
      for(int j=0;j<=m;j++)
    {
      if(a[i]!=a[i-1] && j >=1)//判断该位置的苹果是否连续下降，则进入else；如果苹果不连续，则人可以选择移动或者是不移动。
        {
           if(a[i]==1)//此时如果a[i]处掉苹果时，要分开两种情况进行讨论，你这是可能身处两个地方
        {
        //而你的人这时是可以身处两个位置的，可以位于第一棵树或者第二棵树下。你可以选择移动或者是不移动，这时如果你移动到第一棵树下。此时为位于树1处的最优解
              dp[i][j][1]=max(dp[i-1][j][1],dp[i-1][j-1][2])+1;
       //如果你选择移动到第二棵树下，此时是位于树2处的最优解
        dp[i][j][2]=max(dp[i-1][j][2],dp[i-1][j-1][1]);
        }

      else if(a[i]==2)//第二种情况，类同上面
        {
          dp[i][j][2]=max(dp[i-1][j][2],dp[i-1][j-1][1])+1;
          dp[i][j][1]=max(dp[i-1][j][1],dp[i-1][j-1][2]);
        }
        }

      else //判断这时苹果是连续掉下来的话，此时在两棵树时货的苹果的最优解
        {
          if(a[i]==1)//如果是树1处掉苹果，那么这时你有可能位于两个位置树1和树2
        {
          dp[i][j][1]=dp[i-1][j][1]+1;//当你位于树1时，那么在树1处的最优解就是这棵树的前一秒转移得到的。
          dp[i][j][2]=dp[i-1][j][2]+0;
        }
          else if(a[i]==2)
        {
          dp[i][j][1]=dp[i-1][j][1]+0;
          dp[i][j][2]=dp[i-1][j][2]+1;
        }
        }
    }
    }
      int Max=0;
      for(int i=0;i<=m;i++)
    {
      Max=max(Max,max(dp[n][i][1],dp[n][i][2]));
    }
      printf("%d\n",Max);
  return 0;
}