| Time Limit: 10000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
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
/*
study
zoj2112 树状数组+主席树 区间动第k大
开始想了一下，如果只用主席树在修改的时候差不多要修改所有的树
然后看了下树状数组套主席树，
相当于我们先照静态第k大建好树，然后把修改操作通过树状数组另外建一个树
于是在查询的时候只要通过树状数组得出增减情况再加上最初的
hhh-2016-02-19 15:03:33
*/
#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;
const int maxn = 60010;
int n,m;
int a[maxn],t[maxn];
int T[maxn],val[maxn*40],lson[maxn*40],rson[maxn*40];
int tot;
void ini_hash(int num) //排序去重
{
sort(t,t+num);
m = unique(t,t+num)-t;
}
int Hash(int x) //获得x在排序去重后的位置
{
return lower_bound(t,t+m,x) - t;
}
int build(int l,int r)
{
int root = tot++;
val[root] = 0;
if(l != r)
{
int mid = (l+r)>>1;
lson[root] = build(l,mid);
rson[root] = build(mid+1,r);
}
return root;
}
//如果那里发生改变则兴建一个节点而非像平常修改那个节点的值
int update(int root,int pos,int va)
{
int newroot = tot++;
int tmp = newroot;
val[newroot] = val[root] + va;
int l = 0,r = m-1;
while(l < r)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
lson[newroot] = tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
lson[newroot] = lson[root];
rson[newroot] = tot++;
newroot = rson[newroot];
root = rson[root];
l = mid+1;
}
val[newroot] = val[root] + va;
}
return tmp;
}
int s[maxn];
int cur[maxn];
int lowbit(int x)
{
return x&(-x);
}
//void add(int x,int pos,int va)
//{
// while(x < n)
// {
// s[x] = update(s[x],pos,va);
// x += lowbit(x);
// }
//}
int sum(int x)
{
int cn = 0;
while(x>0)
{
cn += val[lson[cur[x]]];
x -= lowbit(x);
}
return cn;
}
int query(int lt,int rt,int k)
{
int l = 0, r = m-1;
int lroot = T[lt-1];
int rroot = T[rt];
for(int i = lt-1; i; i-=lowbit(i)) cur[i] = s[i];
for(int i = rt; i; i-=lowbit(i)) cur[i] = s[i];
while(l < r)
{
int mid = (l+r)>>1;
int tmp = sum(rt)-sum(lt-1)+val[lson[rroot]]-val[lson[lroot]];
if(tmp >= k)
{
for(int i = lt-1; i; i-=lowbit(i)) cur[i] = lson[cur[i]];
for(int i = rt; i; i-=lowbit(i)) cur[i] = lson[cur[i]];
lroot = lson[lroot];
rroot = lson[rroot];
r = mid;
}
else
{
k -= tmp;
for(int i = lt-1; i; i-=lowbit(i)) cur[i] = rson[cur[i]];
for(int i = rt; i; i-=lowbit(i)) cur[i] = rson[cur[i]];
lroot = rson[lroot];
rroot = rson[rroot];
l = mid+1;
}
}
return l;
}
void change(int root,int pos,int va)
{
while(root <= n)
{
s[root] = update(s[root],pos,va);
root += lowbit(root);
}
}
struct node
{
int l,r,kind,k;
} qry[10010];
int main()
{
int q,cas;
scanf("%d",&cas);
while(cas--)
{
m = 0;
tot = 0;
scanf("%d%d",&n,&q) ;
for(int i = 1; i <= n; i++)
{
scanf("%d",&a[i]);
t[m++] = a[i];
}
char ask[10];
int l,r,k;
for(int i = 1; i <= q; i++)
{
scanf("%s",ask);
if(ask[0] == 'Q')
{
scanf("%d%d%d",&l,&r,&k);
qry[i].l = l;
qry[i].r = r;
qry[i].k = k;
qry[i].kind = 0;
}
else
{
scanf("%d%d",&l,&r);//将l位置的数改为r
qry[i].l = l;
qry[i].r = r;
qry[i].kind = 1;
t[m++] = qry[i].r;
}
}
ini_hash(m);
T[0] = build(0,m-1);
for(int i = 1; i <= n; i++)
T[i] = update(T[i-1],Hash(a[i]),1);
for(int i =1; i <= n; i++)
{
s[i] = T[0];
}
for(int i =1; i <= q; i++)
{
if(qry[i].kind == 0)
{
printf("%d
",t[query(qry[i].l,qry[i].r,qry[i].k)]);
}
else
{
change(qry[i].l,Hash(a[qry[i].l]),-1);
change(qry[i].l,Hash(qry[i].r),1);
a[qry[i].l] = qry[i].r;
}
}
}
return 0;
}