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);
}
欢迎来原博客看看 >原文链接<
BZOJ 1528: [POI2005]sam-Toy Cars、BZOJ 1826: [JSOI2010]缓存交换
原文:https://www.cnblogs.com/Tobichi/p/9084866.html