并查集(UnionSet)

======================= **基础知识** =======================

Union-find 是一种抽象化较高的数据结构，主要用来处理联通性问题，联通性自身具有传递性；

参考文章： 并查集(Union-Find)算法介绍     &   应用举例

 1 //#include "UnionSet.h"
 2 #include <iostream>
 3 using namespace std;
 4 class UnionSet {
 5     public:
 6         int *fa, n;
 7         UnionSet(int n): n(n) {
 8             fa = new int[n + 1];
 9             for(int i = 0 ; i<= n; i++) {
10                 fa[i] = i;
11             }
12         }
13         int find(int x) {
14             return fa[x] = (fa[x] == x ? x : find(fa[x]));
15         }
16         void merge(int a, int b) {
17             fa[find(b)] = find(a);
18             return;
19         }
20 };
21 
22 int main()
23 {
24 //input
25 //6 5
26 //1 1 2
27 //2 1 3
28 //1 2 4
29 //1 4 3
30 //2 1 3
31 
32     int n, m;
33     int a, b, c;
34     scanf("%d %d\n", &n, &m);
35     UnionSet u1(n);
36     while(m--) {
37         int a, b, c;
38         scanf("%d %d %d\n", &a, &b, &c);
39         switch(a) {
40             case 1 :
41                 u1.merge(b, c); break;
42             case 2:{
43                     if(u1.find(b) == u1.find(c)) printf("Yes\n");
44                     else printf("No\n");
45                    } break;;
46             default :
47                 printf("no such op code %d\n", a);
48         }
49     }
50 
51     return 0;
5
1 //#include "UnionSet.h"
 2 #include <iostream>
 3 using namespace std;
 4 class UnionSet {
 5     public:
 6         int *fa, n;
 7         UnionSet(int n): n(n) {
 8             fa = new int[n + 1];
 9             for(int i = 0 ; i<= n; i++) {
10                 fa[i] = i;
11             }
12         }
13         int find(int x) {
14             return fa[x] = (fa[x] == x ? x : find(fa[x]));
15         }
16         void merge(int a, int b) {
17             fa[find(b)] = find(a);
18             return;
19         }
20 };
21 
22 int main()
23 {
24 //input
25 //6 5
26 //1 1 2
27 //2 1 3
28 //1 2 4
29 //1 4 3
30 //2 1 3
31 
32     int n, m;
33     int a, b, c;
34     scanf("%d %d\n", &n, &m);
35     UnionSet u1(n);
36     while(m--) {
37         int a, b, c;
38         scanf("%d %d %d\n", &a, &b, &c);
39         switch(a) {
40             case 1 :
41                 u1.merge(b, c); break;
42             case 2:{
43                     if(u1.find(b) == u1.find(c)) printf("Yes\n");
44                     else printf("No\n");
45                    } break;;
46             default :
47                 printf("no such op code %d\n", a);
48         }
49     }
50 
51     return 0;
52 }
UnionSet

实际使用过程中要具体问题具体分析；

======================= **代码演示** =======================

1. Qucik-Find 算法:  染色法：通过把联通点标定相同颜色；　　查找(联通判断)操作O(1);  合并操作O(n);

　　　　　　　　　　　　    核心思想在于每次merge都会把所有联通点的对应的颜色变为一种颜色，也就是标定哪些点具有相同的颜色(其实也是指向同一个根节点）；

　　　　　　　　　　　　　每次merge 操作都会遍历所有点，这种方式比较呆板，每次都把所有点都过一遍，那就不会漏了；

 1 class UnionSet {
 2     public:
 3         int *color, n;
 4         UnionSet(int n): n(n) {
 5             color = new int[n];
 6         }
 7         ~UnionSet() {
 8             delete color;
 9         }
10         int find(int idx) {
11             if(idx >= n) return -1;
12             return color[idx];
13         }
14         void merge(int a, int b) {
15             if(color[a] == color[b]) return;
16             int cb = color[b];
17             for(int i = 0; i < n; ++i) if(cb == color[i])  color[i] = color[a];
18             return;
19         }
20 };
quick-find

2. Quick-Union 算法：树形结构存储节点，每组root下的包含的节点，是同一个集合； 查找：O(tree-height树高）； 合并操作O(tree-height树高）

　　　　　　　　　　　　　　所以每次要维护的是根节点；在合并过程，如果把树高 大的 合并到树高 小的 ，就会导致树高越来越 大(极限会退化成一个链表(每次都合并到一个单独节点时候));

 1 class UnionSet {
 2     public:
 3         int *boss, n;
 4         UnionSet(int n) : n(n) {
 5             boss = new int[n + 1];
 6             for(int i = 0; i <= n; ++i) boss[i] = i;
 7         }
 8 
 9         int find(int x) {
10             if(boss[x] == x) return x;
11             return find(boss[x]);
12         }
13 
14         void merge(int a, int b) {  //这里是对两个集合合并,并不仅仅是对两个点合并;
15             int f_a = find(a), f_b = find(b); //必须要得到root 节点，然后合并root 节点，这才是合并两个集合，否则的话，值是合并了以b 为root的集合到a；
16             if(f_a == f_b) return;
17             boss[f_b] = f_a; 
18             return;
19         }
20 };
Quick-Union

优化的方式：a. 合并过程中，两个树都遍历到root节点，所以次数没差别；查找过程中，把节点数量 少的root 合并到 节点数量多的 可以减少平均查找次数(被合并的树，所有的节点的查找到root 次数都+1， 所以平均下来，节点少的树被合并能减少平均查找次数);  查找：O(log N);  合并：O（log N）

 1 class UnionSet {
 2     public:
 3         int *boss, *size, n;
 4         UnionSet(int n) : n(n) {
 5             boss = new int[n + 1];
 6             size = new int[n + 1];
 7             for(int i = 0; i <= n; ++i) {
 8                 boss[i] = i;
 9                 size[i] = 1;
10             }
11         }
12 
13         int find(int x) {
14             if(boss[x] == x) return x;
15             return find(boss[x]);
16         }
17 
18         void merge(int a, int b) {  //这里是对两个集合合并,并不仅仅是对两个点合并;
19             int f_a = find(a), f_b = find(b);
20             if(f_a == f_b) return;
21             if(size[f_a] < size[f_b]) {
22                 boss[f_a] = f_b;
23                 size[f_b] += size[f_a];
24             }
25             else {
26                 boss[f_b] = f_a; 
27                 size[f_a] += size[f_b];
28             }
29             return;
30         }
31 };
Quick-Union优化

b. 路径压缩：树高 都变成1，所有节点都挂在root 下面即可；查找：近似O(1);  合并：近似O(1);

 1 class UnionSet {
 2     public:
 3         int *boss, n;
 4         UnionSet(int n) : n(n) {
 5             boss = new int[n + 1];
 6             for(int i = 0; i <= n; ++i) boss[i] = i;
 7         }
 8 
 9         int find(int x) {
10             if(boss[x] == x) return x;
11             return boss[x] = find(boss[x]);
12         }
13 
14         void merge(int a, int b) {  //这里是对两个集合合并,并不仅仅是对两个点合并;
15             int f_a = find(a), f_b = find(b); //必须要得到root 节点，然后合并root 节点，这才是合并两个集合，否则的话，值是合并了以b 为root的集合到a；
16             if(f_a == f_b) return;
17             boss[f_b] = f_a; 
18             return;
19         }
20 };
quickUnion-路径压缩

======================= **经典问题** =======================

1. 简单应用(leetcode200)

 1 class Solution {
 2 public:
 3     class UnionSet{
 4         public:
 5         int n, *father;
 6         UnionSet(int n):n(n) {
 7             father = new int[n + 1];
 8             for(int i = 0; i <= n; ++i) father[i] = i;
 9         }
10         int find(int i) {
11             return father[i] = (father[i] == i ? i : find(father[i]));
12         }
13         void merge(int a, int b) {
14             father[find(b)] = find(a);
15             return;
16         }
17     };
18 
19     int numIslands(vector<vector<char>>& grid) {
20         int m = grid.size(), n = grid[0].size(), ans = 0;
21         int dir[2][2] = {-1, 0, 0, 1};
22         UnionSet u1(m * n);
23         for(int i = 0; i < m; ++i) {  //y
24             for(int j = 0; j < n; ++j) { //x
25                 if('0' == grid[i][j]) continue;
26                 for(int d = 0; d < 2; ++d) {
27                     int nj = j + dir[d][0], ni = i + dir[d][1];
28                     if(nj < 0 || nj >= n || ni < 0 || ni >= m) continue;
29                     if('0' == grid[ni][nj]) continue;
30                     u1.merge(i * n + j, ni * n + nj);
31                 }
32             }
33         }
34 
35         for(int i = 0; i < m; ++i) {
36             for(int j = 0; j < n; ++j) {
37                 int ind = i * n + j;
38                 if('1' == grid[i][j] && u1.find(ind) == ind) ans += 1;
39             }
40         }
41 
42         return ans;
43     }
44 };
UnionSet

2. 相等关系的联通性（leetcode990)

 1 class Solution {
 2 public:
 3     class UnionSet{
 4         public:
 5             int n, *father;
 6             UnionSet(int n): n(n) {
 7                 father = new int[n + 1];
 8                 for(int i = 0; i <= n; ++i) father[i] = i;
 9             }
10             int find(int i) {
11                 return father[i] = (father[i] == i ? i : find(father[i]));
12             }
13             void merge(int a, int b) {
14                 father[find(b)] = find(a);
15                 return;
16             }
17     };
18     bool equationsPossible(vector<string>& equations) {
19         UnionSet u1(26);
20         for(auto &x : equations) {
21             if('!' == x[1]) continue;
22             u1.merge(x[0] - 'a', x[3] - 'a');
23         }
24 
25         for(auto &x : equations) {
26             if('=' == x[1]) continue; 
27             if(u1.find(x[0] - 'a') == u1.find(x[3] - 'a')) return false;
28         }
29         return true;
30     }
31 };
等式的联通性

3. 代表集合关系的树中存在环，且存在所有边的数据，去除不影响集合的边;(leecode684)

 1 class Solution {
 2 public:
 3     struct UnionSet{
 4         int n, *father;
 5         UnionSet(int n): n(n) {
 6             father = new int[n + 1];
 7             for(int i = 0; i <= n; ++i) father[i] = i;
 8         }
 9         int find(int i) {
10             return father[i] = (father[i] == i ? i : find(father[i]));
11         }
12         void merge(int a, int b) {
13             father[find(b)] = find(a);
14             return;
15         }
16     };
17 
18     vector<int> findRedundantConnection(vector<vector<int>>& edges) {
19         int ans = 0, len = edges.size();
20         UnionSet u(len);
21         for(int i = 0; i < len; ++i) {
22             if(u.find(edges[i][0]) == u.find(edges[i][1])) {
23                 ans = i;
24                 continue;
25             }
26             u.merge(edges[i][0], edges[i][1]);
27         }
28         return edges[ans];
29     }
30 };
树中删除多余边

//进阶题：

4. 并查集变种（leetcode128) : a. 存index减少无用数据空间； b. hashmap 快速查找值；c. 并查集变种满足特定需求；

 1 class Solution {
 2 public:
 3     struct UnionSet{
 4         int n, *father, *cnt;
 5         UnionSet(int n): n(n) {
 6             father = new int[n];
 7             cnt = new int[n];
 8             for(int i = 0; i < n; ++i) {
 9                 father[i] = i;
10                 cnt[i] = 1;
11             }
12         }
13 
14         int find(int i) {
15             return father[i] = (father[i] == i ? i : find(father[i]));
16         }
17 
18         void merge(int a, int b) {
19             int fa = find(a), fb = find(b);
20             if(fa == fb) return;
21             father[fb] = fa;
22             cnt[fa] += cnt[fb];
23             return;
24         }
25 
26     };
27     int longestConsecutive(vector<int>& nums) {
28         int len = nums.size();
29         if(0 == len) return 0;
30         UnionSet u1(len);
31         unordered_map<int,int> m1;  //index, val
32 
33         for(int i = 0; i < len; ++i) {
34             if(m1.find(nums[i]) != m1.end()) continue;
35             if(m1.find(nums[i] + 1) != m1.end()) u1.merge(i, m1[nums[i] + 1]);
36             if(m1.find(nums[i] - 1) != m1.end()) u1.merge(i, m1[nums[i] - 1]);
37             m1[nums[i]] = i;
38         }
39 
40         int ans = 0;
41         for(int i = 0; i < len; ++i) {
42             if(u1.find(i) != i) continue; 
43             ans = max(ans, u1.cnt[i]);
44         }
45         return ans;
46 
47 // 排序+ 遍历
48 //        set<int> s1(nums.begin(), nums.end());
49 //        set<int>::iterator x = s1.begin();
50 //        int ans = 0, temp = 1, before = *x;
51 //        while(++x != s1.end()) {
52 //            if((*x - before) == 1) temp += 1;
53 //            else {
54 //                ans = max(ans, temp);
55 //                temp = 1;
56 //            }
57 //            before = *x;
58 //        }
59 //        ans = max(ans, temp);
60 //        return ans;
61     }
62 };
UnionSet变化，cnt 是每个集合中元素数量

  leetcode947  : 这里cnt 存储的就是root 节点数量；

 1 class Solution {
 2 public:
 3     struct UnionSet{
 4         int n, cnt, *father;
 5         UnionSet(int n):n(n), cnt(n) {
 6             father = new int[n];
 7             for(int i = 0;i < n; ++i) father[i] = i;
 8         }
 9         int find(int i) {
10             return father[i] = (father[i] == i ? i : find(father[i]));
11         }
12         void merge(int a, int b) {
13             int fa = find(a), fb = find(b);
14             if(fa == fb) return;
15             father[fb] = fa;
16             cnt -= 1;
17             return;
18         }
19     };
20 
21     int removeStones(vector<vector<int>>& stones) {
22         int len = stones.size();
23         UnionSet u1(len);
24         unordered_map<int, int> mx, my; //val, ind
25         for(int i = 0; i < len; ++i) {
26             int x = stones[i][0], y = stones[i][1];
27             if(mx.find(x) == mx.end()) mx[x] = i;
28             else u1.merge(i, mx[x]);
29 
30             if(my.find(y) == my.end()) my[y] = i;
31             else u1.merge(i, my[y]);
32         }
33 
34         return (len - u1.cnt);
35     }
36 };
UnionSet变化， cnt 是root 节点数量

5. 如何对于各个root 节点下的集合进行操作：

leetcode1202:  对于每个集合中元素进行排列，并输出

 1 class Solution {
 2 public:
 3     struct UnionSet {
 4         int n, *father;
 5         UnionSet(int n):n(n) {
 6             father = new int[n + 1];
 7             for(int i = 0; i <= n; ++i) father[i] = i;
 8         }
 9         int find(int i) {
10             return father[i] = (father[i] == i ? i : find(father[i]));
11         }
12         void merge(int a, int b) {
13             father[find(b)] = find(a);
14             return;
15         }
16     };
17 
18     string smallestStringWithSwaps(string s, vector<vector<int>>& pairs) {
19         UnionSet u1(s.size());
20         for(auto x : pairs) u1.merge(x[0], x[1]);
21 
22         map<int, priority_queue<char,vector<char>, greater<char>>> m;
23         for(int i = 0; s[i]; ++i) m[u1.find(i)].push(s[i]);
24 
25         string ans = "";
26 
27         for(int i = 0; s[i]; ++i) {
28             int idx = u1.find(i);
29             ans += m[idx].top();
30             m[idx].pop();
31         }
32 
33         return ans;
34     }
35 };
UnionSet对每个集合进行排序

 leetcode721

 1 class Solution {
 2 public:
 3     struct UnionSet {
 4         int n, *father;
 5         UnionSet(int n):n(n) {
 6             father = new int[n];
 7             for(int i = 0; i < n; ++i) father[i] = i;
 8         }
 9         int find(int i) {
10             return father[i] = (father[i] == i ? i : find(father[i]));
11         }
12         void merge(int a, int b) {
13             father[find(b)] = find(a);
14             return;
15         }
16     };
17 
18     vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
19         unordered_map<string, int> m2i;
20         int len = accounts.size();
21         UnionSet u1(len);
22         for(int i = 0; i < len; ++i) {
23             for(int j = 1, J = accounts[i].size(); j < J; ++j) {
24                 if(m2i.find(accounts[i][j]) != m2i.end()) u1.merge(i, m2i[accounts[i][j]]);
25                 m2i[accounts[i][j]] = i;
26             }
27         }
28         vector<set<string>> proc(len);
29 
30         for(int i = 0; i < len; ++i) {
31             int idx = u1.find(i);
32             proc[idx].insert(accounts[i].begin() + 1, accounts[i].end());
33         }
34 
35         vector<vector<string>> ans;
36         for(int i = 0; i < len; ++i) {
37             if(u1.find(i) != i) continue;
38             vector<string> temp;
39             temp.push_back(accounts[i][0]);
40             for(auto x : proc[i]) temp.push_back(x);
41             ans.push_back(temp);
42         }
43 
44         return ans;
45     }
UnionSet指定值排序

6. 反复调用UnionSet， 并判断最后一个多余联通条件；leetcode 685

 1 //这里包含了有向图中入度/出度的应用
 2 class Solution {
 3 public:
 4     struct UnionSet{
 5         int n, *father;
 6         UnionSet(int n):n(n) {
 7             father = new int[n + 1];
 8             for(int i = 0; i <= n; ++i) father[i] = i;
 9         }
10         void init(){
11             for(int i = 0; i <= n; ++i) father[i] = i;
12             return;
13         }
14         int find(int i) {
15             return father[i] = (father[i] == i ? i : find(father[i]));
16         }
17         void merge(int a, int b) {
18             father[find(b)] = find(a);
19             return;
20         }
21     };
22 
23     vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
24         int len = edges.size(), ans = len - 1, in_idx = 0, flag = 1;
25         vector<int> in(len + 1, 0);
26         UnionSet u1(len);
27 
28         for(int i = 0; i < len; ++i) {
29             //找入度不正常点(>1)
30             if((++in[edges[i][1]]) < 2) continue;
31             in_idx = edges[i][1];
32             break;
33         }
34 
35         do {
36             //针对入度不正常点的特殊处理
37             if(in_idx && edges[ans][1] != in_idx) continue;
38 
39             //暴力尝试去掉某条边后，就不存在环的情况
40             u1.init();
41             flag = 1;
42             for(int i = len - 1; i >= 0; --i) {
43                 if(ans == i) continue;
44                 if(u1.find(edges[i][0]) == u1.find(edges[i][1])) {
45                     flag = 0;
46                     break;
47                 } else u1.merge(edges[i][0], edges[i][1]);
48             }
49             if(flag) return edges[ans];
50 
51         }while(--ans > 0);
52 
53         return edges[0];
54     }
55 };
反复使用UnionSet

7.1562. 查找大小为 M 的最新分组 : 对于问题和数据处理的转化；

 1 class Solution {
 2     struct UnionSet {
 3         int n;
 4         vector<int> boss, size, cnt;
 5         UnionSet(int _n) : n(_n), boss(_n), size(_n, _n), cnt(_n, 0) {
 6             for(int i = 1; i < _n; ++i) boss[i] = i, size[i] = 0;
 7         }
 8         int find(int a) {
 9             if(boss[a] == a) return a;
10             return boss[a] = find(boss[a]);
11         }
12         void merge(int a, int b) {
13             int fa = find(a), fb = find(b);
14             if(fa == fb) return;
15             boss[fb] = fa;
16             //联通过程，如何计数是关键；
17             //这里设计了一个0点，以及每次联通都将联通后的点链接到前一个点,作为一组;
18             //最终n次操作后，所有的点都链接到0点上；
19             size[cnt[fb]] -= 1;
20             size[cnt[fa]] -= 1;
21             cnt[fa] += (cnt[fb] + 1);
22             size[cnt[fa]] += 1;
23             return;
24         }
25     };
26 public:
27     int findLatestStep(vector<int>& arr, int m) {
28         int ans = -1, len = arr.size();
29         UnionSet us(len + 1);
30 
31         for(int i = 0; i < len; ++i) {
32             us.merge(arr[i] - 1, arr[i]);
33             if(us.size[m])  ans = i + 1;
34         }
35         return ans;
36     }
37 };
问题以及求解的转换

 

======================= **应用场景** =======================

用来处理联通性问题；

树可以看成是一个连通且 无环 的 无向 图。

posted on 2022-01-15 23:16  学海一扁舟  阅读(142)  评论(0)  编辑  收藏  举报
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