划分树水题.....
The Famous ICPC Team Again
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 859 Accepted Submission(s): 415
Problem Description
When Mr. B, Mr. G and Mr. M were preparing for the 2012 ACM-ICPC World Final Contest, Mr. B had collected a large set of contest problems for their daily training. When they decided to take training, Mr. B would choose one of them from the problem set. All
the problems in the problem set had been sorted by their time of publish. Each time Prof. S, their coach, would tell them to choose one problem published within a particular time interval. That is to say, if problems had been sorted in a line, each time they
would choose one of them from a specified segment of the line.
the problems in the problem set had been sorted by their time of publish. Each time Prof. S, their coach, would tell them to choose one problem published within a particular time interval. That is to say, if problems had been sorted in a line, each time they
would choose one of them from a specified segment of the line.
Moreover, when collecting the problems, Mr. B had also known an estimation of each problem’s difficultness. When he was asked to choose a problem, if he chose the easiest one, Mr. G would complain that “Hey, what a trivial problem!”; if he chose the hardest
one, Mr. M would grumble that it took too much time to finish it. To address this dilemma, Mr. B decided to take the one with the medium difficulty. Therefore, he needed a way to know the median number in the given interval of the sequence.
Input
For each test case, the first line contains a single integer n (1 <= n <= 100,000) indicating the total number of problems. The second line contains n integers xi (0 <= xi <= 1,000,000,000), separated by single space, denoting the difficultness of each problem,
already sorted by publish time. The next line contains a single integer m (1 <= m <= 100,000), specifying number of queries. Then m lines follow, each line contains a pair of integers, A and B (1 <= A <= B <= n), denoting that Mr. B needed to choose a problem
between positions A and B (inclusively, positions are counted from 1). It is guaranteed that the number of items between A and B is odd.
already sorted by publish time. The next line contains a single integer m (1 <= m <= 100,000), specifying number of queries. Then m lines follow, each line contains a pair of integers, A and B (1 <= A <= B <= n), denoting that Mr. B needed to choose a problem
between positions A and B (inclusively, positions are counted from 1). It is guaranteed that the number of items between A and B is odd.
Output
For each query, output a single line containing an integer that denotes the difficultness of the problem that Mr. B should choose.
Sample Input
5 5 3 2 4 1 3 1 3 2 4 3 5 5 10 6 4 8 2 3 1 3 2 4 3 5
Sample Output
Case 1: 3 3 2 Case 2: 6 6 4
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100100;
int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];
void build(int l,int r,int dep)
{
if(l==r) return ;
int mid=(l+r)/2;
int same=mid-l+1;
for(int i=l;i<=r;i++)
if(tree[dep][i]<sorted[mid]) same--;
int lpos=l,rpos=mid+1;
for(int i=l;i<=r;i++)
{
if(tree[dep][i]<sorted[mid])
tree[dep+1][lpos++]=tree[dep][i];
else if(tree[dep][i]==sorted[mid]&&same>0)
{
tree[dep+1][lpos++]=tree[dep][i];
same--;
}
else tree[dep+1][rpos++]=tree[dep][i];
toleft[dep][i]=toleft[dep][l-1]+lpos-l;
}
build(l,mid,dep+1);
build(mid+1,r,dep+1);
}
int query(int L,int R,int l,int r,int dep,int k)
{
if(l==r) return tree[dep][l];
int mid=(L+R)/2;
int cnt=toleft[dep][r]-toleft[dep][l-1];
if(cnt>=k)
{
int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
int newr=newl+cnt-1;
return query(L,mid,newl,newr,dep+1,k);
}
else
{
int newr=r+toleft[dep][R]-toleft[dep][r];
int newl=newr-(r-l-cnt);
return query(mid+1,R,newl,newr,dep+1,k-cnt);
}
}
int main()
{
int n,m,cas=1;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",sorted+i);
tree[0][i]=sorted[i];
}
sort(sorted+1,sorted+1+n);
build(1,n,0);
scanf("%d",&m);
printf("Case %d:\n",cas++);
while(m--)
{
int l,r;
scanf("%d%d",&l,&r);
int k=(r-l)/2+1;
printf("%d\n",query(1,n,l,r,0,k));
}
}
return 0;
}