K-th Number hdu 6231（二分+尺取）

最新推荐文章于 2020-10-22 21:07:18 发布

原创最新推荐文章于 2020-10-22 21:07:18 发布 · 197 阅读

1 ·

CC 4.0 BY-SA版权

二分同时被 2 个专栏收录

3 篇文章

订阅专栏

尺取

3 篇文章

订阅专栏

本文介绍了一种算法，用于解决从数组的每个长度大于k的区间中找到该区间第k大的数，并将这些数放入新数组B中，最后求B的第m大元素的问题。通过先对所有数进行排序，然后使用二分查找方法来高效地确定答案。

Alice are given an array A[1..N] with N numbers.

Now Alice want to build an array B by a parameter K as following rules:

Initially, the array B is empty. Consider each interval in array A. If the length of this interval is less than K, then ignore this interval. Otherwise, find the K-th largest number in this interval and add this number into array B.

In fact Alice doesn't care each element in the array B. She only wants to know the M-th largest element in the array B. Please help her to find this number.

Input

The first line is the number of test cases.

For each test case, the first line contains three positive numbers N(1≤N≤105),K(1≤K≤N),M. The second line contains N numbers Ai(1≤Ai≤109).

It's guaranteed that M is not greater than the length of the array B.

Output

For each test case, output a single line containing the M-th largest element in the array B.

Sample Input

5 3 2

2 3 1 5 4

3 3 1

5 8 2

Sampl Output

Source

2017中国大学生程序设计竞赛-哈尔滨站-重现赛（感谢哈理工）

Recommend

jiangzijing2015 | We have carefully selected several similar problems for you: 6447 6446 6445 6444 6443

题意：从每个长度大于k的区间中把这个区间第k大的数放入b中问 b的第m大元素是什么

解：先把所有的数排序，然后二分枚举一个数 x 为答案，查询是否有大于等于 m个区间的第k大值大于 x

#include<stdio.h>
#include<algorithm>
using namespace std;
#define ll long long
ll a[100010],a1[100010];//a为原数组 ,a1为排序数组
ll n,m,k;
ll kk(ll x)//判读这个数是否可能为答案
{
    ll l=1;//起点
    ll en=0;//有多少区间第k大数大于x
    ll sum=0;//区间内有多少个数大于x
    for(ll i=1;i<=n;i++)
    {
        if(a[i]>x) sum++;
        if(sum>=k)//如果这个区间sum>=k 说明这个区间的第k大数 大于 x
        {
            while(sum>=k)
            {
                en+=n-i+1; //以 l 为起点的区间的第k大数肯定都比 x大
                if(a[l++]>x) sum--;//以 l+1为起点
            }
        }
    }
    if(en>=m) return 0;//有超过m个区间，那么答案不可能是 x，肯定比x大
    return 1;//答案可能是x
}
int main()
{
    ll t;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&n,&k,&m);
        for(ll i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            a1[i]=a[i];
        }
        sort(a1+1,a1+1+n);//排序
        ll l=1,r=n,mid,ans;//二分枚举答案
        while(l<=r)
        {
            mid=(l+r)>>1;
            if(kk(a1[mid]))
            {
                ans=mid;
                r=mid-1;
            }
            else
              l=mid+1;
        }
        printf("%lld\n",a1[ans]);
    }
}