我正在解决这个问题：http://www.spoj.com/status/SAM,iiit/
我以某种方式得出了解决方案，但我仍然不能用数学证明它。
问题陈述是什么：
There are 'n' toys (1<=n<=10^5) on a shelf.A child is on the floor.He demands toys in
a sequence to play with , specified by 'p' (1<=p<=5*10^5).His mother gives him a toy
from the shelf if the child demanded a toy which is not on floor.At a time only
'k'(1<=k<=n) toys can be there on floor.So mother when giving toy from shelf can pick a
toy from floor and put it back to shelf if she wants.
So we have to minimize total number of times mother picks toys from shelf.
我的解决方案：
（a）变量和功能：
Keep a set of toys on floor and a variable ans(initially 0),which stores the answer.
Also next[],next[i] tells when will toy number 'i' come next in the demand sequence,
ie. index of its next occurrence in demand sequence.
update next[x] updates next[x] to store the next index of its occurrence in demand
sequence.If there is no further occurrence next[x]=MAX_INTEGER;
（b）算法
Following are the cases:
1.If child demands a 'x' toy from shelf:
increment ans
If there are less than k elements then:
add the element to the set
update next[x]
If there are k elements:
remove the element from set whose value of next[] is largest
add element 'x' to set
update next[x]
2.If child demands toy from floor say toy 'x':
update next[x]
ans is the final answer.
现在我无法证明为什么这种贪婪的方法在数学上是正确的。
答案 0 :(得分：2)
这实际上是一个缓存问题 - 地板是缓存，自我是主存。
您提供的算法是最佳的，因为它只是clairvoyant algorithm。它是一种经典算法，您可以在很多操作系统资源上找到它。