题目地址：见这里
Description

Cluster analysis, or also known as clustering, is a task to group a set of objects into one or more groups
such that objects belong to a same group are more similar compared to object in other groups. In this
problem, you are given a set of N positive integers and an integer K. Your task is to compute how
many clusters are there in the given set where two integers belong to a same cluster if their difference
is no larger than K.
For example, let there be a set of N = 7 positive integers: 2, 6, 1, 7, 3, 4, 9, and K = 1.
Based-on the cluster definition of K, we know that:
• 2 and 1 belong to a same cluster (the difference is no more than K = 1),
• 2 and 3 belong to a same cluster,
• 6 and 7 belong to a same cluster,
• 3 and 4 belong to a same cluster.
From these observations, we can conclude that there are 3 clusters in this example: {2, 1, 3, 4}, {6,
7}, and {9}.
Figure 1.
Figure 1 illustrates the clustering result. A line connecting two numbers means that those two
numbers should belong to a same cluster according to the definition.
Input

The first line of input contains an integer T (T ≤ 100) denoting the number of cases. Each case begins
with two integers in a line: N and K (1 ≤ N ≤ 100; 1 ≤ K ≤ 1, 000, 000) denoting the set size and
the clustering parameter respectively. The next line contains N integers Ai (1 ≤ Ai ≤ 1, 000, 000)
representing the set of positive integers. You are guaranteed that all integers in the set are unique
Output

For each case, output ‘Case #X: Y ’, where X is the case number starts from 1 and Y is the number
of cluster for that particular case.
Explanation for 2nd sample case:
The given set is exactly the same as in 1st sample, however, now K = 2. With two additional
observations (compared to 1st sample): 4 and 6 are in the same cluster, 7 and 9 are in the same cluster;
all those integers will be in the same cluster.
Explanation for 3rd sample case:
There are 2 clusters: {1, 4}, and {15, 20, 17}.
Explanation for 4th sample case:
In this sample, all integers will be in their own cluster.
Sample Input

4
7 1
2 6 1 7 3 4 9
7 2
2 6 1 7 3 4 9
5 5
15 1 20 4 17
8 10
100 200 300 400 500 600 700 800
Sample Output

Case #1: 3
Case #2: 1
Case #3: 2
Case #4: 8

题意就是：有n个数，如果两个数相差小于等于k的话，就可以属于一个集合，问你这n个数，可以属于多少个集合。
解法：数据小，直接暴力枚举+dsu维护

//LA 6906
#include <bits/stdc++.h>
using namespace std;
const int maxn = 115;
int a[maxn], n, k;
namespace dsu{
    int fa[maxn];
    inline void init(){for(int i = 1; i <= n; i++) fa[i] = i; }
    inline int find_set(int x){if(x == fa[x]) return x; else return fa[x] = find_set(fa[x]);}
    inline void union_set(int x, int y){x = find_set(x), y = find_set(y); if(x != y) fa[x] = y;}
}
using namespace dsu;
int main()
{
    int T, ks = 0;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &k);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        init();
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                if(i != j && abs(a[i] - a[j]) <= k) union_set(i, j);
            }
        }
        set <int> s;
        for(int i = 1; i <= n; i++) s.insert(find_set(i));
        printf("Case #%d: %d\n", ++ks, s.size());
    }
    return 0;
}