Educational Codeforces Round 64 (Rated for Div. 2)-C. Match Points

C. Match Points
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

You are given a set of points x1, x2, ..., xn on the number line.

Two points i and j can be matched with each other if the following conditions hold:

neither i nor j is matched with any other point;
|xi−xj|≥z.
What is the maximum number of pairs of points you can match with each other?

Input
The first line contains two integers n and z (2≤n≤2⋅105, 1≤z≤109) — the number of points and the constraint on the distance between matched points, respectively.

The second line contains n integers x1, x2, ..., xn (1≤xi≤109).

Output
Print one integer — the maximum number of pairs of points you can match with each other.

Examples
input
4 2
1 3 3 7
output
2
input
5 5
10 9 5 8 7
output
1
Note
In the first example, you may match point 1 with point 2 (|3−1|≥2), and point 3 with point 4 (|7−3|≥2).

In the second example, you may match point 1 with point 3 (|5−10|≥5).

 题目：http://codeforces.com/contest/1156/problem/C

题意 ： 长度为n的数组， 选择两个数，要求|差值| >= k， 但是被选择过的数不允许被再次选择，问 最大的匹配对数。

在一开始，我想用尺选法+标记数组来求解这个问题，但是卡在了第七个数据上，这组数据的n是200000，正确答案是100000，而我输出了9913，小于正确答案。

我就很困惑，直到看到别人博客里的一组数据  n = 4,k = 4  arr =  {4 8 9 12} 这组数据用我一开始的程序来算的话是 1 而正确答案是2

尺选法+标记数组 WA on test7

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 2e5+5;
int arr[maxn];
int vis[maxn]={};
int main(){
    int n, k;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
        scanf("%d",&arr[i]);
    sort(arr,arr+n);
    int cnt = 0;
    int i=0,j=0;
    while(j<n){
        while(arr[j]-arr[i] < k && j<n)j++;
        if(j>=n)break;
        vis[j] = 1;
        j++;
        cnt++;
        i++;
        while(vis[i] && i<n)i++;
    }
    printf("%d
",cnt);
    return 0;
}

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 2e5+5;
int arr[maxn];

int main(){
    int n, k;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
        scanf("%d",&arr[i]);
    sort(arr,arr+n);
    int cnt = 0;
    int i=0,j=n/2;
    for(;i<n/2;i++){
        for(;j<n;j++){
            if(arr[j]-arr[i]>=k){
                cnt++;j++;
                break;
            }
        }
    }
    printf("%d
",cnt);
    return 0;
}

还可以二分做，

链接 : https://blog.csdn.net/pythonbanana/article/details/89786072