Points on the line
题面：
We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1, 3, 2, 1} is 2.
Diameter of multiset consisting of one point is 0.
You are given n points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed d?
The first line contains two integers n and d (1 ≤ n ≤ 100, 0 ≤ d ≤ 100) — the amount of points and the maximum allowed diameter respectively.
The second line contains n space separated integers (1 ≤ xi ≤ 100) — the coordinates of the points.
Output a single integer — the minimum number of points you have to remove.
3 1 2 1 4
1
3 0 7 7 7
0
6 3 1 3 4 6 9 10
3
In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.
In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.
看上去好像是搜索，想了一会儿后发现既然是最大值和最小值之间的差，那么数组有序后，就是取一个中间的区间。但是同样是符合条件的区间，用什么办法能取到所求的最大区间？其实不需要，数据很小直接暴力就行了。其实想到是取一个区间这个题就已经做出来了。
- #include <cstdio>
- #include <cstring>
- #include <iostream>
- #include <cmath>
- #include <map>
- #include <algorithm>
- #include <vector>
- #define INF 0x3f3f3f3f
- using namespace std;
- int arr[105];
- int main()
- {
- int n,d;
- cin>>n>>d;
- for(int i=0;i<n;i++)
- cin>>arr[i];
- sort(arr,arr+n);
- int ans=-INF;
- for(int i=0;i<n;i++)
- {
- for(int j=0;j<n;j++)
- {
- if(arr[j]-arr[i]<=d)
- ans=max(ans,j-i+1);
- }
- }
- cout<<n-ans<<endl;
- return 0;
- }