poj 2104 K-th Number (分桶法和平方分割)

K-th Number

Time Limit: 20000MS Memory Limit: 65536KTotal Submissions: 53887 Accepted: 18530Case Time Limit: 2000MS

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment. 
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?" 
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000). 
The second line contains n different integer numbers not exceeding 10⁹ by their absolute values --- the array for which the answers should be given. 
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 31 5 2 6 3 7 42 5 34 4 11 7 3

Sample Output

563

题意：给出一个数列，询问其中一段区间 [L,R]中第k大的数

题解： m的数很大所以直接每次都快排的方法不可取，看到网上有很多人用的是主席树和线段树……然而都不会……

资料书上正好用这道题进行讲解的 分桶法和平方分割 的数据结构 ，看了好长时间，大概给懂了

分桶法：是把一排物品或者平面分成桶，每个桶分别维护自己内部的信息，以达到高效的计算的目的的方法。

其中，平方分割是把排成一排的n个元素每 根号下n个分在一个桶里进行维护的方法的统称。将复杂度降到sqrt（n）。

如果x是第k个数，那么一定有：在区间中不超过x的数有不少于k个；  在区间中小于x的数有不到k个

通过分桶法，每个桶中保留B个有序的数据，被区间完全包含的桶直接二分搜索小于等于k的数的个数，不被完全包含的桶就逐个遍历

code

#include <iostream>#include<cstdio>#include<algorithm>#include<vector>using namespace std;#define MAXN 100010#define MAXM 5010#define B 1000   //桶的大小//输入int n,m;int a[MAXN];int L[MAXM],R[MAXM],K[MAXM];int num[MAXN];  //对A排序后的结果vector<int> bucket[MAXN/B];  //每个桶排序后的结果void solve(){    for(int i=0;i<n;i++)    {        bucket[i/B].push_back(a[i]);        num[i]=a[i];    }    sort(num,num+n);    for(int i=0;i<n/B;i++)        sort(bucket[i].begin(),bucket[i].end());    for(int i=0;i<m;i++)    {        //求[L,R]区间的第K个数        int l=L[i]-1,r=R[i],k=K[i];        int left=-1,right=n-1,mid;        while(right-left>1)        {            mid=(left+right)/2;            int x=num[mid];            int tl=l,tr=r,c=0;            //区间两端多出的部分            while(tl<tr&&tl%B!=0){                if(a[tl++]<=x) c++;            }            while(tl<tr&&tr%B!=0){                if(a[--tr]<=x) c++;            }            //对每个桶进行计算            while(tl<tr)            {                int  b=tl/B;                c+=upper_bound(bucket[b].begin(),bucket[b].end(),x)-bucket[b].begin();                tl+=B;            }            if(c>=k)                right=mid;            else                left=mid;        }        printf("%d\n",num[right]);    }}int main(){    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)        scanf("%d",&a[i]);    for(int i=0;i<m;i++)        scanf("%d%d%d",&L[i],&R[i],&K[i]);    solve();    return 0;}