POJ 2385 Apple Catching（牛吃苹果）

Apple Catching

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 12817
Accepted: 6227

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

题目大意：有两棵树，每分钟其中的一棵树会掉苹果下来，牛一开始再树1下，最多可以移动W次，问牛最多可以吃到多少苹果。

解题思路：

1、牛最开始在树1下

2、吃到最多的苹果不一定是移动了W次

3、每一分钟牛所在的位置 = 牛移动的次数%2 + 1

4、初始化方程为 dp[ i ][ 0 ]=dp[ i - 1 ][ 0 ]+ a[ i ]%2;

5、动规方程为dp[ i ][ j ]=max(dp[ i - 1 ][ j ],dp[ i - 1 ][ j - 1 ]);

如果在这分钟有苹果掉到牛所在的位置，责dp[ i ][ j ]++

AC代码

#include <stdio.h>  #include <stdlib.h>  #include <string.h>  int arr[1005];  int dp[1005][35];  int main()  {      int t,w;      while (scanf("%d %d", &t, &w) != EOF)      {          memset(dp, 0, sizeof(dp));          for (int i = 1; i <= t; ++ i )          {              scanf("%d", &arr[i]);          }          //初始化，开始在第一颗树下          if (arr[1] == 1)          {              dp[1][0] = 1;              dp[1][1] = 0;          }          if (arr[1] == 2)          {              dp[1][1] = 1;              dp[1][0] = 0;          }            for (int i = 2; i <= t; ++ i)          {              for (int j = 0; j <= w; ++ j)              {                  //初始化                  if (j == 0)                  {                      dp[i][j] = dp[i - 1][j] + arr[i] % 2;                      continue;                  }                  //dp状态方程                  dp[i][j] = dp[i - 1][j] > dp[i - 1][j - 1] ? dp[i - 1][j] : dp[i - 1][j - 1];                  //如果本次是在第i颗树下，就会多收获一个苹果                  if (j % 2 + 1 == arr[i])                  {                      dp[i][j] ++;                  }              }          }          //找最大值          int ans = dp[t][0];          for (int i = 0; i <= w ; ++ i)          {              if (ans < dp[t][i])              {                  ans = dp[t][i];              }          }          printf("%d\n", ans);      }      return 0;  }