POJ 2385 -- Apple Catching _Apple

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Apple Catching

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13234		Accepted: 6437

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.【POJ 2385 -- Apple Catching】Sample Input

7 2 2 1 1 2 2 1 1

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

文章图片

1 #include < iostream> 2 #include < cmath> 3 #include < cstring> 4 #include < cstdio> 5 #include < cstdlib> 6 #include < algorithm> 7 using namespace std; 8 int a[1010]; 9 int f[1010]; 10 int dp[1010][1010]; 11 int main() 12 { 13int T,W; 14scanf("%d%d",& T,& W); 15for(int i=1; i< =T; i++) 16scanf("%d",& a[i]); 17int pre=0; 18for(int i=1; i< =T; i++)//先把相邻的相同的苹果合在一起，这样相邻的i,i+1就是指不同树上的苹果了 19{ 20if(a[i]==a[i-1]) f[pre]++; 21else {pre++; f[pre]=1; } 22} 23for(int i=1; i< =pre; i++) 24{ 25dp[i][0]=f[i]; //init：如果不动，能吃到当前位置上的果子 26for(int j=1; j< =W; j++) 27{ 28dp[i][j]=max(dp[i-2][j],dp[i-1][j-1])+f[i]; //此时相邻的是不同树，中间隔一个的是同一棵树上的 29} 30} 31int ans=0; 32for(int i=0; i< =W; i++)//此处注意：不一定步数都用完就能吃到最多的 33ans=max(ans,dp[pre][i]); 34printf("%d",ans); 35system("pause"); 36return 0; 37 }

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