1 second
256 megabytes
standard input
standard output
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.
感想：本题逆向思考比较适合，之前的想法还是没过样例，后面的代码是错误的，先放着。现在写模拟题，最怕WA了但是不知道WA哪里，这种情况下，怎么才能快速找到bug或者怎么从一种思维模式中挣脱出来尝试新方法呢...
code
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
int main()
{
int n,d;
cin>>n>>d;
int a[];
for(int i=;i<n;i++)
cin>>a[i];
sort(a,a+n);
int ans=;
for(int i=;i<n;i++)
{
int pos=i,t=;
while(a[pos]<=a[i]+d&&pos<n) //默认a[i]前的数都已经被删除了
{
pos++;
t++;
} ans=max(ans,t);
}
cout<<n-ans;
}
wrong code（错了两次，还没改正）
#include<string.h>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
int main()
{
int n,d;
int a[];
int b[];
while(~scanf("%d%d",&n,&d))
{
int ans=;
memset(a,,sizeof(a));
memset(b,,sizeof(b));
for(int i=;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
for(int i=;i<n;i++)
for(int j=;j<n;j++)
if(abs(a[i]-a[j])>d)
b[a[i]]++;
int r=n-,l=;
while(a[r]-a[l]>d)
{
if(b[a[r]]>b[a[l]])
{
r--;
for(int i=;i<n;i++)
if(abs(a[r]-a[i])>d) //
b[a[i]]--;
}
else
{
l++;
for(int i=;i<n;i++)
if(abs(a[l]-a[i])>d) //这两个地方是新添的
b[a[i]]--;
}
ans++;
}
printf("%d\n",ans); }
}