终于开始写dp了,还很不熟练
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
* Lines 2..T+1: 1 or 2: the tree that will drop an apple each minute.
Output
Sample Input
7 2
2
1
1
2
2
1
1
Sample Output
6
Hint
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.
状态转移方程:dp[i][j] = max(dp[i-1][j], dp[i-1][j-1]),然后判断当前是否在第i分钟掉苹果的那颗树下,是的话,dp[i][j]++。
对状态转移方程的解释如下:第i分钟能得到的苹果数量,等于在第i-1分钟时,在树1和树2下得到苹果的最大值。j为偶数则在树1下面,奇数则在树2下面。
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