UVa 11987 Almost Union-Find（支持删除操作的并查集）

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Description

I hope you know the beautiful Union-Find structure. In this problem, you’re to implement something similar, but not identical. The data structure you need to write is also a collection of disjoint sets, supporting 3 operations:

• 1 p q Union the sets containing p and q. If p and q are already in the same set, ignore this command.
• 2 p q Move p to the set containing q. If p and q are already in the same set, ignore this command.
• 3 p Return the number of elements and the sum of elements in the set containing p.

Initially, the collection contains n sets: {1}, {2}, {3}, . . . , {n}.

Input

There are several test cases. Each test case begins with a line containing two integers n and m (1 ≤ n, m ≤ 100, 000), the number of integers, and the number of commands. Each of the next m lines contains a command. For every operation, 1 ≤ p, q ≤ n. The input is terminated by end-of-file (EOF).

Output

For each type-3 command, output 2 integers: the number of elements and the sum of elements.

Explanation

Initially: {1}, {2}, {3}, {4}, {5}

Collection after operation 1 1 2: {1,2}, {3}, {4}, {5}

Collection after operation 2 3 4: {1,2}, {3,4}, {5} (we omit the empty set that is produced when taking out 3 from {3})

Collection after operation 1 3 5: {1,2}, {3,4,5}

Collection after operation 2 4 1: {1,2,4}, {3,5}

Sample Input

5 7 
1 1 2 
2 3 4 
1 3 5 
3 4 
2 4 1 
3 4 
3 3

Sample Output

3 12 
3 7 
2 8

思路

题意：

给出1-N的数，一开始每个数自形成一个集合，要求支持以下操作

• 将元素 p 所在集合与元素 q 所在集合合并
• 将元素 p 移到元素 q 所在集合
• 询问元素 p 所在集合有多少个元素，和为多少

题解:

第一和第三个操作都是很裸的并查集，关键在于操作2，很容易可以想到，元素 p 移到元素 q所在集合，可以先在元素 p 所在集合将其删除，然后把元素 p 与元素 q 所在集合合并。而这一步的关键在于删除操作，如果 p 是叶子节点，直接改变 p 的父亲指向就行，但是 p 即所在集合的根节点又当如何，将其删除的话其子节点需要重新建立关系，这个过程还是有一定复杂性的。我们可以采取另外的思路：二次哈希法（ReHash），对于每个结点都有一个哈希值，在进行查找之前需要将x转化成它的哈希值HASH[x]，那么在进行删除的时候，只要将x的哈希值进行改变，变成一个从来没有出现过的值（可以采用一个计数器来实现这一步），然后对新的值建立集合，因为只有它一个元素，所以必定是一个新的集合。这样做可以保证每次删除操作的时间复杂度都是O(1)的，而且不会破坏原有树结构，唯一的一个缺点就是每删除一个结点其实是多申请了一块内存，如果删除操作无限制，那么内存会无限增长。

简单的说，就是建立虚拟节点，这样原来的节点还在树当中，并不会破坏其结构，开辟一个id[ ]数组表示元素 x 对应的值。

 

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#include<bits/stdc++.h> using namespace std; const int maxn = 100005; int cnt,fa[maxn],num[maxn],sum[maxn],id[maxn];   void init(int N) {     for (int i = 0;i <= N;i++)     {         fa[i] = id[i] = sum[i] = i;         num[i] = 1;     }     cnt = N; }   int find(int x) {     int r = x;     while (r != fa[r])  r = fa[r];     int i = x,j;     while (i != r)     {         j = fa[i];         fa[i] = r;         i = j;     }     return r; }   void Union(int x,int y) {     int fx = find(id[x]),fy = find(id[y]);     fa[fx] = fy;     num[fy] += num[fx];     sum[fy] += sum[fx]; }   void Delete(int x) {     int fx = find(id[x]);     --num[fx];     sum[fx] -= x;     id[x] = ++cnt,fa[id[x]] = id[x],num[id[x]] = 1,sum[id[x]] = x; }   int main() {     int N,M;     while (~scanf("%d%d",&N,&M))     {         int opt,x,y;         init(N);         while (M--)         {             scanf("%d",&opt);             if (opt == 1)             {                 scanf("%d%d",&x,&y);                 if (find(id[x]) != find(id[y])) Union(x,y);             }             else if (opt == 2)             {                 scanf("%d%d",&x,&y);                 if (find(id[x]) != find(id[y])) Delete(x),Union(x,y);             }             else             {                 scanf("%d",&x);                 printf("%d %d\n",num[find(id[x])],sum[find(id[x])]);             }         }     }     return 0; }