HDU2665-Kth number

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最新推荐文章于 2024-05-09 00:45:37 发布

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分类专栏： HDU ----线段树

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本文介绍了一种解决区间内查找第K大数的问题的方法。通过预先排序和重新编号，利用可持久化线段树进行高效维护和查询。适用于大规模数据集，提供了一个完整的C++实现示例。

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Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9899 Accepted Submission(s): 3070

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

Input

The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

Output

For each test case, output m lines. Each line contains the kth big number.

Sample Input

  
   1 
10 1 
1 4 2 3 5 6 7 8 9 0 
1 3 2

Sample Output

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

题意：给出n个数，有m个询问，求区间内第k大的数是哪个

解题思路：由于题目未说明n个数的范围，因此可以先做一个预处理，将n个数排序后重新编号，然后建一棵可持久化线段树来维护

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <bitset>
#include <set>
#include <vector>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, q,m;
int a[100009], b[100009],tot;
int L[100009 * 20], R[100009 * 20], s[100009], sum[100009 * 20];

void build(int pre,int &now, int l, int r, int k)
{
	sum[now=++tot] = sum[pre] + 1;
	if (l == r) { L[now] = R[now] = 0; return; }
	int mid = (l + r) >> 1;
	L[now] = L[pre], R[now] = R[pre];
	if (mid >= k) build(L[pre],L[now], l, mid, k);
	else build(R[pre],R[now], mid + 1, r, k);
}

int query(int k, int kk, int l, int r, int p)
{
	if (l == r) return l;
	int mid = (l + r) >> 1;
	if (p > sum[L[k]] - sum[L[kk]]) return query(R[k], R[kk], mid + 1, r, p - sum[L[k]] + sum[L[kk]]);
	else return query(L[k], L[kk], l, mid, p);
}

int main()
{
	while (~scanf("%d%d", &n,&q))
	{
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
		sort(b + 1, b + 1 + n);
		m = unique(b + 1, b + 1 + n) - b;
		L[0] = R[0] = sum[0] = tot =s[0]= 0;
		for (int i = 1; i <= n; i++)
		{
			a[i] = lower_bound(b + 1, b + m, a[i]) - b;
			build(s[i - 1],s[i], 1, m, a[i]);
		}
		while (q--)
		{
			int l, r, k;
			scanf("%d%d%d", &l, &r, &k);
			printf("%d\n", b[query(s[r], s[l-1], 1, m, k)]);
		}
	}
	return 0;
}