bzoj1901&zoj2112&cogs257 Dynamic Rankings(动态排名系统)


cogs

zoj

bzoj-权限


题解

bzoj和zoj都是骗访问量的233,我没有权限

带修改区间k小值,看了学习了题解一次AC真开心。。。

不带修改的是前缀和套主席树,\(O(log_2n)\)查询,要修改只能\(O(nlog_2n)\)乱搞。

把外层前缀和改成树状数组,即第\(i\)个主席树从存\(1\) ~ \(i\)的值域改成存\(i-lowbit(i)+1\) ~ \(i\)的值域。然后修改和查询都是\(O(log_2^2n)\)的。。。(一个\(log\)树状数组另一个线段树)

查询就从两个根相减变成多个根减多个根了,和bit差不多。


Code

// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "dynrank"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
#define lb(o) ((o)&(-(o)))
#define mid ((l+r)>>1)
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
typedef struct node* point;
point null;
struct node{
int data;
point ls,rs;
node(){ls=rs=null,data=0;}
node(point _ls,point _rs){ls=_ls,rs=_rs,data=0;}
};
const int maxn=50001,maxm=10001;
il vd copy(point&a,point b){
if(b==null)a=null;
else a=new node,*a=*b;
}
il vd Updata(point&s,point rt,int l,int r,int&pos,int num){
copy(s,rt);
s->data+=num;
if(l==r)return;
if(mid<pos)Updata(s->rs,rt->rs,mid+1,r,pos,num);
else Updata(s->ls,rt->ls,l,mid,pos,num);
}
il point build(int l,int r){
if(l==r)return new node;
return new node(build(l,mid),build(mid+1,r));
}
point a[18],b[18];
int _a,_b;
il int Query(int l,int r,int k){
while(l<r){
int res=0;
rep(i,1,_a)res+=a[i]->ls->data;
rep(i,1,_b)res-=b[i]->ls->data;
if(res>=k){
rep(i,1,_a)a[i]=a[i]->ls;
rep(i,1,_b)b[i]=b[i]->ls;
r=mid;
}else{
rep(i,1,_a)a[i]=a[i]->rs;
rep(i,1,_b)b[i]=b[i]->rs;
l=mid+1,k-=res;
}
}return l;
}
il vd free(point i){if(i!=null)free(i->ls),free(i->rs),delete i;}
il vd work(){
int n=gi(),m=gi(),N=n;
point root[maxn+maxm];
int num[maxn],data[maxn+maxm];
int A[maxm],B[maxm],K[maxm];
char opt[3];
rep(i,1,n)num[i]=data[i]=gi();
rep(i,1,m){
scanf("%s",opt);
A[i]=gi(),B[i]=gi();
if(opt[0]=='Q')K[i]=gi();
else K[i]=-1,data[++N]=B[i];
}
sort(data+1,data+N+1);
int tot=unique(data+1,data+N+1)-data-1;
root[0]=build(1,tot);
rep(i,1,n)num[i]=lower_bound(data+1,data+tot+1,num[i])-data,root[i]=root[i-1];
rep(i,1,n)for(rg int j=i;j<=n;j+=lb(j))Updata(root[j],root[j],1,tot,num[i],1);
rep(i,1,m){
if(K[i]==-1){
B[i]=lower_bound(data+1,data+tot+1,B[i])-data;
for(rg int j=A[i];j<=n;j+=lb(j))Updata(root[j],root[j],1,tot,num[A[i]],-1);
num[A[i]]=B[i];
for(rg int j=A[i];j<=n;j+=lb(j))Updata(root[j],root[j],1,tot,num[A[i]],1);
}else{
_a=_b=0;
for(rg int j=B[i];j;j-=lb(j))a[++_a]=root[j];
for(rg int j=A[i]-1;j;j-=lb(j))b[++_b]=root[j];
printf("%d\n",data[Query(1,tot,K[i])]);
}
}
}
int main(){
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
int T=gi();
null=new node;
null->ls=null->rs=null;
while(T--)work();
return 0;
}
04-21 09:39