[poj 2104 K-th Number] 主席树 区间第K大

[poj 2104 K-th Number] 主席树 区间第K大

题目链接：[poj 2104 K-th Number]

题意描述：给定N个数a1,a2,…,an，M次查询，每次查询区间第L个数到第R个数中的第K大数。

相似题目： [hdu 4417 Super Mario] 主席树+离散化

解题思路：首先对数据离散化。然后线段树记录区间中的数字出现的次数。主席树保存N棵线段树，每棵线段树都保存着前N个数字的信息。查询区间[L,R]的第K大， 只要根据第L−1棵数和第R棵树二分向叶子节点就好了。

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define fst             first
#define snd             second
#define __mid__         int mid = ((l + r) >> 1)

typedef __int64 LL;
typedef unsigned __int64 ULL;
typedef pair<int, int> PII;

const int MAXN = 100000 + 5;
const int MAXM = 5000 + 5;
const int INF = 0x3f3f3f3f;

int N, M, A[MAXN], F[MAXN], L, R, K;
int root[MAXN], FSZ, TSZ;

struct TNode {
int ls, rs, sum;
} node[MAXN * 20];
void hash_init() {
sort(F, F + N);
FSZ = unique(F, F + N) - F;
F[FSZ ++] = INF;
}
int getID(const int& x) { return lower_bound(F, F + FSZ, x) - F + 1;}
int build(int l, int r) {
int rt = TSZ ++;
node[rt].sum = 0;
if(l == r) { node[rt].ls = node[rt].rs = -1; return rt;}
__mid__;
node[rt].ls = build(l, mid);
node[rt].rs = build(mid + 1, r);
return rt;
}
int update(const int& pos, int r1, int l, int r) {
int r2 = TSZ ++;
node[r2] = node[r1];
node[r2].sum ++;
if(l == r) return r2;
__mid__;
if(pos <= mid) node[r2].ls = update(pos, node[r1].ls, l, mid);
else node[r2].rs = update(pos, node[r1].rs, mid + 1, r);
return r2;
}
int query(int k, int r1, int r2, int l, int r) {
if(l == r) return l;
__mid__;
int x = node[node[r2].ls].sum - node[node[r1].ls].sum;
if(k <= x) return query(k, node[r1].ls, node[r2].ls, l, mid);
else return query(k - x, node[r1].rs, node[r2].rs, mid + 1, r);
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
while(~scanf("%d %d", &N, &M)) {
for(int i = 0; i < N; i++) scanf("%d", &A[i]), F[i] = A[i];
hash_init();
TSZ = 0;
root[0] = build(1, FSZ);
for(int i = 0; i < N; i++) {
root[i + 1] = update(getID(A[i]), root[i], 1, FSZ);
}
while(M --) {
scanf("%d %d %d", &L, &R, &K);
int ret = query(K, root[L - 1], root[R], 1, FSZ) - 1;
printf("%d\n", F[ret]);
}
}
return 0;
}