并不对劲的bzoj1861: [Zjoi2006]Book 书架

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这题的正确做法是splay维护这摞书。

但是并不对劲的人选择了暴力（皮这一下很开心）。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define re register
#define maxn 300010
using namespace std;
inline int read()
{
    int x=0,f=1;
    char ch=getchar();
    while(isdigit(ch)==0 && ch!='-')ch=getchar();
    if(ch=='-')f=-1,ch=getchar();
    while(isdigit(ch))x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
inline void write(int x)
{
    int f=0;char ch[20];
    if(!x){puts("0");return;}
    if(x<0){putchar('-');x=-x;}
    while(x)ch[++f]=x%10+'0',x/=10;
    while(f)putchar(ch[f--]);
    putchar('
');
}
int a[maxn],tr[maxn],pos[maxn],n,m,qx,qy,hd,tl;
char s[20];
int lt(int x){return x&(-x);}
void add(int x,int k){for(;x<=n+m*2;x+=lt(x))tr[x]+=k;}
int ask(int x){int k=0;for(;x;x-=lt(x))k+=tr[x];return k;}
int kth(int k)
{
    int l=hd,r=tl,ans=tl+233;
    while(l<=r)
    {
        int mi=(r+l)>>1;
        int rank=ask(mi);
        if(rank==k)ans=min(ans,mi);
        if(rank>=k)r=mi-1;
        else l=mi+1;
    }
    return ans;
}
int main()
{
    n=read(),m=read();hd=1+m,tl=n+m;
    rep(i,1,n)a[i+m]=read(),pos[a[i+m]]=i+m,add(i+m,1);
    rep(k,1,m)
    {
    	scanf("%s",s);qx=read();
    	if(s[0]=='T')add(pos[qx],-1),hd--,add(hd,1),swap(a[hd],a[pos[qx]]),pos[qx]=hd;
    	if(s[0]=='B')add(pos[qx],-1),tl++,add(tl,1),swap(a[tl],a[pos[qx]]),pos[qx]=tl;
    	if(s[0]=='I')
    	{
    		qy=read();if(!qy)continue;
    		int tmp=ask(pos[qx])+qy;
    	//	cout<<"rank:"<<tmp<<endl;
    		int ans=kth(tmp);
    	//	cout<<"ans&pos:"<<qx<<" "<<a[ans]<<pos[qx]<<" "; 
            int px=pos[qx];
            swap(pos[qx],pos[a[ans]]);swap(a[px],a[ans]);// cout<<pos[qx]<<endl;
    	}
    	if(s[0]=='A')write(ask(pos[qx])-1);
    	if(s[0]=='Q')
    	{
    		int ans=kth(qx);
    		write(a[ans]);
    	}
   	//cout<<"a:"<<endl;
   	//rep(i,1,n+m*2)cout<<a[i]<<" ";cout<<endl;
   	//cout<<"pos:"<<endl;
   	//rep(i,1,n)cout<<pos[i]<<" ";cout<<endl;
    }
    return 0;
}
/*
10 10
1 3 2 7 5 8 10 4 9 6
Query 3
Top 5
Ask 6
Bottom 3
Ask 3
Top 6
Insert 4 -1
Query 5
Query 2
Ask 2
*/

　　

据说照顾了splay的大肠腧的数据为暴力提供了漏洞。