codeforces日记371e

最新推荐文章于 2017-07-10 17:34:29 发布

原创最新推荐文章于 2017-07-10 17:34:29 发布 · 562 阅读

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本文介绍了一种算法，用于从现有地铁站中选择最优的k个站点以最小化乘客平均通勤时间。通过排序和计算不同站点组合间的距离，该算法能够有效地找到最佳方案。

题目：

Berland is going through tough times — the dirt price has dropped and that is a blow to the country's economy. Everybody knows that Berland is the top world dirt exporter!

The President of Berland was forced to leave only k of the currently existing n subway stations.

The subway stations are located on a straight line one after another, the trains consecutively visit the stations as they move. You can assume that the stations are on the Ox axis, the i-th station is at point with coordinate xi. In such case the distance between stations iand j is calculated by a simple formula |xi - xj|.

Currently, the Ministry of Transport is choosing which stations to close and which ones to leave. Obviously, the residents of the capital won't be too enthusiastic about the innovation, so it was decided to show the best side to the people. The Ministry of Transport wants to choose such k stations that minimize the average commute time in the subway!

Assuming that the train speed is constant (it is a fixed value), the average commute time in the subway is calculated as the sum of pairwise distances between stations, divided by the number of pairs (that is $\text{[math]}$ ) and divided by the speed of the train.

Help the Minister of Transport to solve this difficult problem. Write a program that, given the location of the stations selects such kstations that the average commute time in the subway is minimized.

Input

The first line of the input contains integer n (3 ≤ n ≤ 3·105) — the number of the stations before the innovation. The second line contains the coordinates of the stations x1, x2, ..., xn ( - 108 ≤ xi ≤ 108). The third line contains integer k (2 ≤ k ≤ n - 1) — the number of stations after the innovation.

The station coordinates are distinct and not necessarily sorted.

Output

Print a sequence of k distinct integers t1, t2, ..., tk (1 ≤ tj ≤ n) — the numbers of the stations that should be left after the innovation in arbitrary order. Assume that the stations are numbered 1 through n in the order they are given in the input. The number of stations you print must have the minimum possible average commute time among all possible ways to choose k stations. If there are multiple such ways, you are allowed to print any of them.

Sample test(s)

input
3
1 100 101
2

output
2 3 

Note

In the sample testcase the optimal answer is to destroy the first station (with x = 1). The average commute time will be equal to 1 in this way.

集体思路：

将输入的车站以及位置用pair保存，按位置进行排序。可知所求的必为排序后连续的k个车站，计算前k个车站所有对的距离之和（后发现不用计算即代码中注释掉的，只用看距离的大小即可），根据前一个计算后推一个车站的距离之和，记录最小值并输出。

AC代码：

#include<iostream>
#include<algorithm>
using namespace std;
pair<int, int> x[300008];
long long S[300008],i,n,k,m,mi,f;

int main() {
 for (cin>>n; i<n; i++) {
  cin >> x[i].first;
  x[i].second = i+1;
 }
 cin>>k; k--;
 sort(x, x+n);
 for (i=1; i<n; i++) S[i] = S[i-1] + x[i].first;
 f=0;
 //for (i=1; i<=k; i++) f += x[i].first * i - S[i-1];
 for (m=f,i=1; i<n-k; i++)
  if ((f += x[i+k].first*k-2*S[i+k-1]+2*S[i-1]+x[i-1].first*k) < m) m=f,mi=i;
 for (i=0; i<=k; i++)
  cout << x[mi+i].second << " ";
}