Dynamic Rankings
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2112
Description
The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.
Your task is to write a program for this computer, which
- Reads N numbers from the input (1 <= N <= 50,000)
- Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change some a[i] to t.
Input
The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.
The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format
Q i j k or
C i t
It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.
There‘re NO breakline between two continuous test cases.
Output
For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],..., a[j])
There‘re NO breakline between two continuous test cases.
Sample Input
2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
Sample Output
3
6
3
6
HINT
题意
动态第k大
1. Q x y k,查询区间[x,y]第k大的数是啥
2. C x y，把第x个数变成y
题解：
树套树
在每一个区间里面，我们都套入一个Treap/Splay就好了
1操作很简单，暴力进行二分就好了，我们在[L,R]区间所在的Treap里面查询Rank
2操作，在所有这个数所在的区间，我们都删除原来的，再加入新的就好了
代码：
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
#include<ctime>
using namespace std;
#define maxn 1000001
int tmp = 0;
////////////////treap
struct Treap
{
int root[maxn],sz,s[maxn],ls[maxn],rs[maxn],v[maxn],w[maxn],rnd[maxn];
void init()
{
memset(root,0,sizeof(root));
sz=0;
}
void Updata(int k)
{
s[k]=s[ls[k]]+s[rs[k]]+w[k];
}
void Rturn(int &k)
{
int t;
t=ls[k],ls[k]=rs[t],rs[t]=k,s[t]=s[k];
Updata(k);k=t;
}
void Lturn(int &k)
{
int t;
t=rs[k],rs[k]=ls[t],ls[t]=k,s[t]=s[k];
Updata(k);k=t;
}
void Insert(int &k,int num)
{
if(!k)
{
k=++sz;s[k]=w[k]=1;ls[k]=rs[k]=0;rnd[k]=rand();
v[k]=num;return;
}
s[k]++;
if(v[k]==num)w[k]++;
else if(num<v[k])
{
Insert(ls[k],num);
if(rnd[ls[k]]<rnd[k])
Rturn(k);
}
else
{
Insert(rs[k],num);
if(rnd[rs[k]]<rnd[k])
Lturn(k);
}
}
void Del(int &k,int num)
{
if(v[k]==num)
{
if(w[k]>1){
w[k]--;
s[k]--;
return;
}
if(ls[k]*rs[k]==0)
k=ls[k]+rs[k];
else if(rnd[ls[k]]<rnd[rs[k]])
Rturn(k),Del(k,num);
else
Lturn(k),Del(k,num);
}
else if(num<v[k]){
Del(ls[k],num);
s[k]--;
}
else{
Del(rs[k],num);
s[k]--;
}
}
void Find(int k,int num)
{
if(!k)return;
if(v[k]<=num){
tmp+=s[ls[k]]+w[k];
Find(rs[k],num);
}
else Find(ls[k],num);
}
}Tr;
/////////////////////
/////////////////////线段树
void Seg_insert(int k,int l,int r,int x,int num)
{
Tr.Insert(Tr.root[k],num);
if(l==r)return;
int mid=(l+r)/2;
if(x<=mid)Seg_insert(k*2,l,mid,x,num);
else Seg_insert(k*2+1,mid+1,r,x,num);
}
void Seg_change(int k,int l,int r,int x,int Now,int Pre)
{
Tr.Del(Tr.root[k],Pre);
Tr.Insert(Tr.root[k],Now);
if(l==r)return;
int mid=(l+r)/2;
if(x<=mid)Seg_change(k*2,l,mid,x,Now,Pre);
else Seg_change(k*2+1,mid+1,r,x,Now,Pre);
}
void Seg_query(int k,int l,int r,int L,int R,int num)
{
if(l==L&&r==R)
{
Tr.Find(Tr.root[k],num);
return;
}
int mid = (l+r)/2;
if(mid>=R)
Seg_query(k*2,l,mid,L,R,num);
else if(mid<L)
Seg_query(k*2+1,mid+1,r,L,R,num);
else{
Seg_query(k*2,l,mid,L,mid,num);
Seg_query(k*2+1,mid+1,r,mid+1,R,num);
}
}
///////////////////////////
int a[maxn];
int main()
{
srand(time(NULL));
int t;scanf("%d",&t);
while(t--)
{
Tr.init();
int n,m;scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
Seg_insert(1,1,n,i,a[i]);
}
char op[10];
for(int i=1;i<=m;i++)
{
scanf("%s",op);
if(op[0]==‘C‘)
{
int x,y;scanf("%d%d",&x,&y);
Seg_change(1,1,n,x,y,a[x]);
a[x]=y;
}
else
{
int x,y,z;scanf("%d%d%d",&x,&y,&z);
int l = 0,r = 1e9;
while(l<=r)
{
int mid = (l+r)/2;
tmp = 0;Seg_query(1,1,n,x,y,mid);
if(tmp>=z)r=mid-1;
else l=mid+1;
}
printf("%d\n",l);
}
}
}
}