4
贪心,每当要把玩具放回去的时候放一个距离当前最远的,正确性显然,1826只需要对a数组进行离散化,思路一样;
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <iostream> #include <cctype> using namespace std; inline char nc() { static char buf[100000], *p1, *p2; return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; } inline int read() { int x=0;char ch=nc(); while(!isdigit(ch))ch=nc(); while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-‘0‘;ch=nc();} return x; } struct node { int next,x; bool operator < (const node &x) const { return next<x.next; } }; int head[510000],a[510000],next[510000]; bool vis[510000]; priority_queue<node>q; int main() { /*freopen("toy.in","r",stdin); freopen("toy.out","w",stdout);*/ int n,k,p; n=read(),k=read(),p=read(); for(int i=1;i<=p;i++)a[i]=read(); for(int i=1;i<=n;i++)head[i]=p+1; for(int i=1;i<=p;i++) { next[head[a[i]]]=i; head[a[i]]=i; } int now=0,ans=0; for(int i=1;i<=p;i++) if(!next[i])next[i]=p+1; for(int i=1;i<=p;i++) { if(vis[a[i]]) { node tmp; tmp.x=a[i]; tmp.next=next[i]; q.push(tmp); } else if(now<k) { now++; node tmp; tmp.x=a[i]; tmp.next=next[i]; q.push(tmp); vis[a[i]]=1; ans++; } else { node tmp=q.top(); while(!vis[tmp.x]) { //q.pop(); tmp=q.top(); } q.pop(); vis[tmp.x]=0; tmp.x=a[i]; tmp.next=next[i]; ans++; q.push(tmp); vis[a[i]]=1; } } printf("%d\n",ans); }
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BZOJ 1528: [POI2005]sam-Toy Cars、BZOJ 1826: [JSOI2010]缓存交换
原文:https://www.cnblogs.com/Tobichi/p/9084866.html