hdu 4251 The Famous ICPC Team Again划分树入门题

The Famous ICPC Team Again

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1091    Accepted Submission(s): 530

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.

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.

 

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

给定n个数

q次询问，每次要求输出询问区间中的中间值

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

const int MAXN=100010;
int tree[30][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序的数
int toleft[30][MAXN];//toleft[p][i]表示第i层从1到i有多少个数分入左边

void build(int l,int r,int dep)
{
    if(l==r)return;
    int mid=(l+r)>>1;
    int same=mid-l+1;//表示等于中间值而且被分入左边的个数
    for(int i=l;i<=r;i++)
      if(tree[dep][i]<sorted[mid])
         same--;
    int lpos=l;
    int rpos=mid+1;
    for(int i=l;i<=r;i++)
    {
        if(tree[dep][i]<sorted[mid])//比中间的数小，分入左边
             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  //比中间值大分入右边
            tree[dep+1][rpos++]=tree[dep][i];
        toleft[dep][i]=toleft[dep][l-1]+lpos-l;//从1到i放左边的个数

    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);

}


//查询区间第k大的数，[L,R]是大区间，[l,r]是要查询的小区间
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)>>1;
    int cnt=toleft[dep][r]-toleft[dep][l-1];//[l,r]中位于左边的个数
    if(cnt>=k)
    {
        //L+要查询的区间前被放在左边的个数
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
        //左端点加上查询区间会被放在左边的个数
        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 T;
    int n,m;
    int s,t,k;
    int cnt=0;
    while(scanf("%d",&n)!=EOF)
    {
        cnt++;
        //scanf("%d%d",&n,&m);
        memset(tree,0,sizeof(tree));//这个必须
        for(int i=1;i<=n;i++)//从1开始
        {
            scanf("%d",&tree[0][i]);
            sorted[i]=tree[0][i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);
        scanf("%d",&m);
        printf("Case %d:
",cnt);
        while(m--)
        {
            scanf("%d%d",&s,&t);
            k=1+(t-s)/2;//此处即为欲求的中间值属于第几大
            printf("%d
",query(1,n,s,t,0,k));
        }
    }
    return 0;
}