[POJ 2104] K-th Number【莫队+树状数组】
Problem:
| Time Limit: 20000MS | Memory Limit: 65536K | |
| 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.
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).
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.
Source
Northeastern Europe 2004, Northern Subregion
Solution:
这道题我之前用归并树做过，参考 [POJ 2104] K-th Number【归并树】。
今天学了莫队算法，据说可以“处理一切区间问题”，就用莫队做了一下 ~~
这次的莫队要和树状数组分工合作。
首先要离散化。
在更新的同时，维护区间 [l, r] 中 ≤ i 的数的个数。
询问第 k 小时，我们在树状数组上二分下标 i，询问 ≤ i 的数的个数，直到 i == k。
关于莫队详见: [BZOJ 2038][国家集训队2009]小Z的袜子【莫队】。
Code: O(nlogn+n1.5+mn0.5+mlogm+mlog2n) [2332K, 2485MS]
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
inline void getint(int &num){
char ch; int sgn = 1;
while(!isdigit(ch = getchar())) if(ch == '-') sgn = -1;
num = ch - '0';
while(isdigit(ch = getchar())) num = (num << 1) + (num << 3) + ch - '0';
num *= sgn;
}
int n, m, a[100005], bl[100005];
int disc[100005], ans[5005];
struct Query{
int id, l, r, k;
} q[5005];
inline bool cmp(const Query &q1, const Query &q2){
if(bl[q1.l] == bl[q2.l]) return q1.r < q2.r;
return q1.l < q2.l;
}
struct BIT{
int node[100005];
inline void add(int u, int v) {while(u < n) node[u] += v, u += u & -u;}
inline int query(int u) {int res = 0; while(u) res += node[u], u -= u & -u; return res;}
inline int inv_rank(int k){
int l = 1, r = n;
while(l < r){
int mid = l + r >> 1;
if(query(mid) >= k) r = mid;
else l = mid + 1;
}
return l;
}
} b;
inline void update(int pos, int val) {b.add(a[pos], val);}
inline void Mo(){
sort(q + 1, q + m + 1, cmp);
for(register int i = 1, l = 1, r = 0; i <= m; i++){
while(r < q[i].r) update(++r, 1);
while(r > q[i].r) update(r--, -1);
while(l < q[i].l) update(l++, -1);
while(l > q[i].l) update(--l, 1);
ans[q[i].id] = disc[b.inv_rank(q[i].k)]; // Must take the original value
}
}
int main(){
getint(n), getint(m);
int blocksize = int(sqrt(n));
for(register int i = 1; i <= n; i++) bl[i] = (i - 1) / blocksize + 1;
for(register int i = 1; i <= n; i++) getint(a[i]), disc[i] = a[i];
for(register int i = 1; i <= m; i++)
getint(q[i].l), getint(q[i].r), getint(q[i].k), q[i].id = i;
sort(disc + 1, disc + n + 1);
int dsize = unique(disc + 1, disc + n + 1) - disc - 1;
for(register int i = 1; i <= n; i++)
a[i] = lower_bound(disc + 1, disc + dsize + 1, a[i]) - disc;
// Discretization
Mo();
for(register int i = 1; i <= m; i++) printf("%d\n", ans[i]);
return 0;
}
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