题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3415
题目大意:给出一个有N个数字(N<=10^5)的环状序列,让你求一个和最大的连续子序列。这个连续子序列的长度小于等于K。
分析:因为序列是环状的,所以可以在序列后面复制前k-1个数字。如果用s[i]来表示复制过后的序列的前i个数的和,那么任意一个子序列[i..j]的和就等于s[j]-s[i-1]。对于每一个j,用s[j]减去最小的一个s[i](i>=j-k)就可以得到以j为终点长度不大于k的和最大的序列了。将原问题转化为这样一个问题后,就可以用单调队列解决了。
单调队列即保持队列中的元素单调递增(或递减)的这样一个队列,可以从两头删除,只能从队尾插入。单调队列的具体作用在于,由于保持队列中的元素满足单调性,对于上述问题中的每个j,可以用O(1)的时间找到对应的s[i]。(保持队列中的元素单调递增的话,队首元素便是所要的元素了)。
维护方法:对于每个j,我们插入s[j-1]的下标,插入时从队尾插入。为了保证队列的单调性,我们从队尾开始删除元素,直到队尾元素对应的值比当前需要插入的s[j-1]小,就将当前元素下标插入到队尾。之所以可以将之前的队列尾部元素全部删除,是因为它们已经不可能成为最优的元素了,因为当前要插入的元素位置比它们靠前,对应的值比它们小。我们要找的,是满足(i>=j-k)的i中最小的s[i]。在插入元素后,从队首开始,将不符合限制条件(i<j-k)的元素全部删除,此时队列一定不为空。(因为刚刚插入了一个一定符合条件的元素)。详细见代码:
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