poj 2104 K-th Number(函数式线段树)

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K-th Number

Time Limit: 20000MS Memory Limit: 65536KTotal Submissions: 31500 Accepted: 9704Case 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

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

Source

Northeastern Europe 2004, Northern Subregion

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 100010#define mid(i,j)  ((i+j)>>1)struct node{int l,r,cnt;void init(int ll,int rr,int s){l=ll;r=rr;cnt=s;}}T[N*20];int a[N],b[N],r[N],n,m,cnt;void build(int l,int r,int &k){k==cnt++;T[k].init(0,0,0);if(l==r)return;int mid=mid(l,r);build(l,mid,T[k].l);build(mid+1,r,T[k].r);}void update(int p,int x,int l,int r,int &k){k=cnt++;T[k].init(T[p].l,T[p].r,T[p].cnt+1);if(l==r)return;int mid=mid(l,r);if(mid>=x)update(T[p].l,x,l,mid,T[k].l);//住左else update(T[p].r,x,mid+1,r,T[k].r);//往右}int query(int i,int j,int l,int r,int k){//[i,j]中第k大if(l==r)return r;int mid=mid(l,r);int t=T[T[j].l].cnt-T[T[i].l].cnt;if(t>=k)return query(T[i].l,T[j].l,l,mid,k);else return query(T[i].r,T[j].r,mid+1,r,k-t);}int main(){int i,j,k;while(scanf("%d%d",&n,&m)!=EOF){cnt=1;for(i=1;i<=n;i++){scanf("%d",&a[i]);b[i]=a[i];}sort(b+1,b+n+1);//从小到大排int num=unique(b+1,b+n+1)-b-1;//除相邻重复元素后的元素个数build(1,num,r[0]); for(i=1;i<=n;i++){a[i]=lower_bound(b+1,b+num+1,a[i])-b;//返回第一个大于a[i]的位置update(r[i-1],a[i],1,num,r[i]);}while(m--){scanf("%d%d%d",&i,&j,&k);printf("%d\n",b[query(r[i-1],r[j],1,num,k)]);}}return 0;}