hdu 4251 The Famous ICPC Team Again（划分树）

The Famous ICPC Team Again

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1101    Accepted Submission(s): 538

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

55 3 2 4 131 32 43 5510 6 4 8 231 32 43 5

 

Sample Output

Case 1:332Case 2:664

 

给定n个值 求区间中的中位数 

也是求区间中第k大的值 k的大小与给定的区间左右端点的大小有关

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 100010#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int tree[20][MAXN];//表示每层每个位置的值int sorted[MAXN];//已经排序好的数int toleft[20][MAXN];//toleft[p][i]表示第i层从1到i有数分到左边void build(int l,int r,int dep){    if(l==r) return;    int mid=(l+r)/2;    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])            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);}//查询区间第K大的数 [L,R]是大区间 [l,r]是要查询的小空间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)/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(){//    fread;    int cs=1;    int n,m;    while(scanf("%d",&n)!=EOF)    {        MEM(tree,0);        for(int i=1;i<=n;i++)        {            scanf("%d",&tree[0][i]);            sorted[i]=tree[0][i];        }        sort(sorted+1,sorted+n+1);        build(1,n,0);        int l,r,k;        scanf("%d",&m);        printf("Case %d:\n",cs++);        while(m--)        {            scanf("%d%d",&l,&r);            k=(r-l)/2+1;            printf("%d\n",query(1,n,l,r,0,k));        }    }    return 0;}