hdu 5412 CRB and Queries 2015多校联合训练赛#10 分治 求区间第k大数
CRB and Queries
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 636 Accepted Submission(s): 162
Problem Description
There are
N boys in CodeLand.
Boy i has his coding skill Ai .
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 l v
The coding skill of Boy l has changed to v .
Query 2: 2 l r k
This is a report query which asks the k -th smallest value of coding skill between Boy l and Boy r (both inclusive).
Boy i has his coding skill Ai .
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 l v
The coding skill of Boy l has changed to v .
Query 2: 2 l r k
This is a report query which asks the k -th smallest value of coding skill between Boy l and Boy r (both inclusive).
Input
There are multiple test cases.
The first line contains a single integer N .
Next line contains N space separated integers A1 , A2 , …, AN , where Ai denotes initial coding skill of Boy i .
Next line contains a single integer Q representing the number of queries.
Next Q lines contain queries which can be any of the two types.
1 ≤ N , Q ≤ 105
1 ≤ Ai , v ≤ 109
1 ≤ l ≤ r ≤ N
1 ≤ k ≤ r – l + 1
The first line contains a single integer N .
Next line contains N space separated integers A1 , A2 , …, AN , where Ai denotes initial coding skill of Boy i .
Next line contains a single integer Q representing the number of queries.
Next Q lines contain queries which can be any of the two types.
1 ≤ N , Q ≤ 105
1 ≤ Ai , v ≤ 109
1 ≤ l ≤ r ≤ N
1 ≤ k ≤ r – l + 1
Output
For each query of type 2, output a single integer corresponding to the answer in a single line.
Sample Input
5 1 2 3 4 5 3 2 2 4 2 1 3 6 2 2 4 2
Sample Output
3 4
Author
KUT（DPRK）
Source
2015 Multi-University Training Contest 10
求区间第k大数，http://www.cnblogs.com/zig-zag/archive/2013/04/18/3027707.html 参考这篇博客，然后看2013年集训队XHR论文。吧
解法：操作分三种，
1. 加入一个数，
2. 删除一个数，
3. 询问答案在左区间还是又区间
方法：
二分询问的答案是什么。
对于初始数字，变为插入操作
按操作的时间顺序排列各个操作，对于修改操作拆为删除和加入操作
：1 删除之前插入的数字，2. 加入新的数字
接下来分治二分答案：
对于mid，如果插入或者删除的数字<=mid那么应该放到左区间，并且用树状数组（其他数据结构也行）
维护前X个位置有多少个数字在左边。
对于询问：如果l,r区间在左边的数字>=k那么答案在左边，否则答案在右边，并且更新K，k=k-这个区间去左边的数字个数
当low =high的时候，说明low就是答案了，只要更新询问还在low，low之间的答案即可。
复杂度分析：分治的深度是log（S）s是数据的范围。
每个询问每次被分到左边或者右边，会有log(s)次操作。每个插入或者删除也是。
但是要用树状数组维护多少个数字在左边，要log(n)次更新或者查询。
复杂度是n*log(s)*log（n)
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
#define maxn 300007
int tree[maxn];
void add(int p,int n){
for(;p<maxn;p+=p&(-p))
tree[p]+=n;
}
int query(int p){
int ans = 0;
for(;p>0;p-=p&(-p))
ans += tree[p];
return ans;
}
struct Node{
int l,r,k,ty,ans;
};
Node p[maxn];
int id1[maxn],id2[maxn];
void CDQ(int L,int R,int low,int high){
if(R < L) return ;
if(low == high ){
for(;L<=R;L++){
p[id1[L]].ans = low;
}
return ;
}
int mid = (low+high)/2,l=L,r=R,k,u;
for(int i = L;i <= R; i++){
u = id1[i];
if(p[u].ty == 2){
k = query(p[u].r) - query(p[u].l-1);
if(k >= p[u].k) id2[l++] = u;
else {
p[u].k -= k;
id2[r--] = u;
}
}
else if(p[u].k <= mid){
add(p[u].l,p[u].ty);
id2[l++] = u;
}
else id2[r--] = u;
}
for(int i = L; i <= R; i++){
u = id1[i];
if(p[u].ty != 2 && p[u].k <= mid) add(p[u].l,-p[u].ty);
}
for(k=L;k<l;k++)
id1[k] = id2[k];
for(r=R;k<=R;k++)
id1[k] = id2[r--];
CDQ(L,l-1,low,mid);
CDQ(l,R,mid+1,high);
}
int num[maxn];
int main(){
int n,q,t,cnt;
memset(tree,0,sizeof(tree));
while(scanf("%d",&n)!=EOF){
for(cnt=0;cnt<n;cnt++){
scanf("%d",&p[cnt].k);
p[cnt].ty = 1;
p[cnt].l = cnt+1;
num[cnt+1] = p[cnt].k;
}
scanf("%d",&q);
int ty,l,v;
for(int i = 0;i < q; i++,cnt++){
scanf("%d",&p[cnt].ty);
if(p[cnt].ty == 1){
scanf("%d%d",&l,&v);
p[cnt].ty = -1;
p[cnt].k = num[l];
p[cnt].l = l;
cnt++;
num[l] = v;
p[cnt].ty = 1;
p[cnt].k = v;
p[cnt].l = l;
}
else {
scanf("%d%d%d",&p[cnt].l,&p[cnt].r,&p[cnt].k);
}
}
for(int i = 0;i < cnt; i++)
id1[i] = i;
CDQ(0,cnt-1,0,1000000000);
for(int i = 0;i < cnt; i++){
if(p[i].ty == 2) printf("%d\n",p[i].ans);
}
}
return 0;
}