HDU 2665 Kth number (主席树)

题意：给定一个序列，求给定区间的第 k 小的值。

析：就是一个主席树的裸板。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn], b[maxn];
int ls[maxn*20], rs[maxn*20], tot, c[maxn*20];
int root[maxn];

void build(int l, int r, int &rt){
  rt = tot++;
  c[rt] = 0;
  if(l == r)  return ;
  int m = l+r >> 1;
  build(l, m, ls[rt]);
  build(m+1, r, rs[rt]);
}

void update(int o, int p, int v, int l, int r, int &rt){
  rt = tot++;
  c[rt] = c[o] + v;
  ls[rt] = ls[o];
  rs[rt] = rs[o];
  if(l == r)  return ;
  int m = l+r >> 1;
  if(p <= m)  update(ls[o], p, v, l, m, ls[rt]);
  else  update(rs[o], p, v, m+1, r, rs[rt]);
}

int query(int o, int k, int l, int r, int rt){
  if(l == r)  return l;
  int m = l+r >> 1;
  int res = c[ls[rt]] - c[ls[o]];
  if(res >= k)  return query(ls[o], k, l, m, ls[rt]);
  return query(rs[o], k-res, m+1, r, rs[rt]);
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d", &n, &m);
    tot = 0;
    for(int i = 0; i < n; ++i){
      scanf("%d", a+i);
      b[i] = a[i];
    }
    sort(b, b + n);
    int sz = unique(b, b + n) - b;
    build(1, sz, root[0]);
    for(int i = 0; i < n; ++i){
      int j = lower_bound(b, b + sz, a[i]) - b + 1;
      update(root[i], j, 1, 1, sz, root[i+1]);
    }
    while(m--){
      int l, r, v;
      scanf("%d %d %d", &l, &r, &v);
      printf("%d\n", b[query(root[l-1], v, 1, sz, root[r])-1]);
    }
  }
  return 0;
}