【POJ 2104 K-th Number】+ 主席树

K-th Number
Time Limit: 20000MS Memory Limit: 65536K
Total Submissions: 51458 Accepted: 17601
Case 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 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

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

本来在学平面桶分割法~却各种T~最后发现需要用主席树~=.=~几乎裸主席树~一次建树一次查询~（划分树和函数式线段树也是能写的~）

AC代码：

#include<cstdio>#include<algorithm>using namespace std;const int MAXN = 1e5 + 10;struct node1{    int L,R,sum;}st[MAXN * 20];struct node2{    int x,idx;    bool operator < (const node2 &px) const{    return x < px.x;    }}a[MAXN];int root[MAXN],vis[MAXN],N,M,nl,fl,fr,fk;void ins(int &num,int &x,int L,int R){     st[++nl] = st[x]; x = nl; ++st[x].sum;     if(L == R) return ;     int mid = (L + R) >> 1;     if(num <= mid) ins(num,st[x].L,L,mid);     else ins(num,st[x].R,mid + 1,R);}int que(int i,int j,int k,int L,int R){     if(L == R) return L;     int t = st[st[j].L].sum - st[st[i].L].sum;     int mid = (L + R) >> 1;     if(k <= t) return que(st[i].L,st[j].L,k,L,mid);     else return que(st[i].R,st[j].R,k - t,mid + 1 , R);}int main(){    st[0].L = st[0].R = st[0].sum = root[0] = 0;    scanf("%d %d",&N,&M);    for(int i = 1 ; i <= N; i++){        scanf("%d",&a[i].x);        a[i].idx = i;    }    sort(a + 1 , a + 1 + N);    for(int i = 1 ; i <= N; i++) vis[a[i].idx] = i;    nl = 0;    for(int i = 1 ; i <= N; i++){        root[i] = root[i - 1];        ins(vis[i],root[i],1,N);    }    while(M--){        scanf("%d %d %d",&fl,&fr,&fk);        printf("%d\n",a[que(root[fl - 1],root[fr],fk,1,N)].x);    }    return 0;}