题目描述
Johnny is a little boy - he is only three years old and enjoys playing with toy cars very much. Johnny has nnn different cars. They are kept on a shelf so high, that Johnny cannot reach it by himself. As there is little space in his room, at no moment may there be more than kkk toy cars on the floor.
Johnny plays with one of the cars on the floor. Johnny's mother remains in the room with her son all the time. When Johnny wants to play with another car that is on the floor, he reaches it by himself. But when the toy is on the shelf, his mummy has to hand it to him. When she gives Johnny one car, she can at the same time pick any car from the floor and put it back on the shelf (so that there remains sufficient space on the floor).
The mother knows her child very well and therefore can predict perfectly which cars Johnny will want to play with. Having this knowledge, she wants to minimize the number of times she has to hand Johnny a toy from the shelf. Keeping that in mind, she has to put the toys off on the shelf extremely thoughtfully.
TaskWrite a programme that:
reads from the standard input the sequence of toy cars in order in which Johnny will want to play with them,calculates the minimal number of times the mother has to pick cars from the shelf,writes the result to the standard output.
(Jasio) 是一个三岁的小男孩,他最喜欢玩玩具了,他有(n)个不同的玩具,它们都被放在了很高的架子上所以(Jasio) 拿不到它们. 为了让他的房间有足够的空间,在任何时刻地板上都不会有超过(k) 个玩具. (Jasio) 在地板上玩玩具. (Jasio)'的妈妈则在房间里陪他的儿子. 当(Jasio) 想玩地板上的其他玩具时,他会自己去拿,如果他想玩的玩具在架子上,他的妈妈则会帮他去拿,当她拿玩具的时候,顺便也会将一个地板上的玩具放上架子使得地板上有足够的空间. 他的妈妈很清楚自己的孩子所以他能够预料到(Jasio) 想玩些什么玩具. 所以她想尽量的使自己去架子上拿玩具的次数尽量的少,应该怎么安排放玩具的顺序呢?
输入格式
In the first line of the standard input there are three integers: (n,k,p) ((1le kle nle 100 000,1le ple 500 000)), separated by single spaces. These denote respectively: the total number of cars, the number of cars that can remain on the floor at once and the length of the sequence of cars which Johnny will want to play with. Each of the following ppp lines contains one integer. These integers are the numbers of cars Johnny will want to play with (the cars are numbered from (1) to (n)).
输出格式
In the first and only line of the standard output one integer should be written - the minimal number of times his mother has to pick a car from the shelf.
输入输出样例
输入
3 2 7
1
2
3
1
3
1
2
输出
4
分析
这个题的主要思路就是贪心，具体思路楼下说的应该挺清楚了，这里再进行一番具体的分析（魔改）
思路
题意是我们要求出来最小的步数能让(Jasio)把所有想玩的玩完。那么我们就可以跟饿狼一样贪了。
考虑一下，假如说地上不足(k)个物品并且当前地上没有(Jasio)想要玩的玩具，那么肯定就直接取下来就行。
假如地上已经满足了(k)个东西了，那么这时候是需要放回一个物品，所以放回哪个物品就是这个题的关键。
因为地上的东西已经有(k)个了，而从当前向后可能还会有要玩某个玩具的需求，那么我们就可以找出当前地上所有东西之中，下一次玩最靠后的那个玩具拿走就行，而这个下一次玩最靠后的玩具可以用大根堆来进行维护，在前边反着枚举预处理。
因为大根堆（优先队列）无法进行单独的删除操作，每次只能把堆顶取出，但是如果之前入过队的，但是和当前需要入队的玩具是同种，那么直接取出是不行的，但是当前这个玩具的下一次玩肯定是比上一个要靠后，那么我们就可以把(k++)，让上一个这种玩具永远保留在里边，为了记录这个玩具出队，我们可以使用一个(vis)数组，每次堆中元素个数等于(k)了，就让堆顶元素的(vis)清空，然后弹出堆顶，然后统计答案并将当前要玩的东西加入队列即可。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
const int maxm = 5e5+10;
int jl[maxm],a[maxm];
int vis[maxn],pos[maxn];
priority_queue<pair<int,int> >q;//按pair第一个元素大根堆排序
int main(){
int n,k,p;
scanf("%d%d%d",&n,&k,&p);
for(int i=1;i<=p;++i){
scanf("%d",&a[i]);
}
for(int i=p;i>=1;--i){//反着枚举
if(!pos[a[i]])jl[i] = 0x3f3f3f3f;//后边不出现了就极大值，不影响
else jl[i] = pos[a[i]];//记录下一次当前玩具出现的时候
pos[a[i]] = i;//记录出现位置
}
int ans = 0;
for(int i=1;i<=p;++i){
if(vis[a[i]]){//如果之前取过但是没出队，就保留着
k++;//保留了一个就让限制++
q.push(make_pair(jl[i],a[i]));//入队
}
else{
if(q.size() == k){//地上放不下了，就出队
vis[q.top().second] = 0;//出队了说明没取，vis数组清除
q.pop();
}
ans++;//答案++
q.push(make_pair(jl[i],a[i]));//入队
vis[a[i]] = 1;//记录当前玩具入队了
}
}
printf("%d
",ans);
}