分析
贪心策略很好想，显然，当 \(cache\) 未满时，直接放进去就行了； \(cache\) 满了的时候，考虑交换哪一个，不难看出可以交换下一次出现最晚的那个。
注意：不开 \(map\) 过不了！
代码
#include <map>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define il inline
#define re register
#define maxn 100005
#define tie0 cin.tie(0),cout.tie(0)
#define fastio ios::sync_with_stdio(false)
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
template <typename T> inline void read(T &x) {
T f = 1; x = 0; char c;
for (c = getchar(); !isdigit(c); c = getchar()) if (c == '-') f = -1;
for ( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
x *= f;
}
priority_queue <P> q;
map <int, int> lst, inq;
int n, m, ans;
int a[maxn], nxt[maxn];
int main() {
memset(nxt, 1, sizeof(nxt));
read(n), read(m);
for (int i = 1; i <= n; ++i) {
read(a[i]);
nxt[lst[a[i]]] = i;
lst[a[i]] = i;
}
for (int i = 1; i <= n; ++i) {
if (inq[a[i]]) {
q.push(make_pair(nxt[i], a[i]));
m++;
continue;
}
if (q.size() < m) {
q.push(make_pair(nxt[i], a[i]));
ans++; inq[a[i]] = 1;
continue;
}
int k = q.top().second; q.pop();
inq[k] = 0;
q.push(make_pair(nxt[i], a[i]));
inq[a[i]] = 1;
ans++;
}
printf("%d", ans);
return 0;
}