codeforces802 A&&B Heidi and Library (easy&&medium)
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题目大意:
每天会有一个客人来图书馆,共n天,第i个客人要看的书是ai。
为了保证每个客人都能读到想要的书,你可以在任意时间去商店买书并放入图书馆或者从图书馆里丢掉几本书,图书馆里最多只能有k本书。问最少买多少本书?
枚举每一天,如果ai已经在图书馆里就不用买了。如果不在,若书的总数小于k,就买这本书,否则丢掉下次出现的时间最晚的书。显然这样一定是最优的。
代码:
#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<vector>#include<queue> using namespace std;#define N 400010#define INF 100000000struct Node{ int w,f; Node(){} Node(int w,int f):w(w),f(f){} bool operator < (Node a)const{ return w<a.w; }}Tmp;priority_queue<Node>Q;int i,j,k,s,Ans,n,m,p[N],a[N],M,x,y,nxt[N],l[N];bool b[N];int main(){ scanf("%d%d",&n,&k); for(i=1;i<=n;i++)scanf("%d",&a[i]),l[a[i]]=INF; for(i=n;i>=1;i--) nxt[i]=l[a[i]],l[a[i]]=i; for(i=1;i<=n;i++) if(!b[a[i]]){ if(s==k){ b[Q.top().f]=0;s--;Q.pop(); } Ans++;Q.push(Node(nxt[i],a[i]));b[a[i]]=1;s++; }else Q.push(Node(nxt[i],a[i])); printf("%d\n",Ans); return 0;}
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