UVA - 11987 Almost Union-Find（带删除的并查集）

Almost Union-Find

Time Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

Submit Status

Description

Problem A

Almost Union-Find

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). The size of input file does not exceed 5MB.

Output

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

Sample Input

5 71 1 22 3 41 3 53 42 4 13 43 3

Output for the Sample Input

3 123 72 8

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}

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Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!

 

带删除的并查集.

操作2实际上是将p从原来的集合中删除，再加入到q所在的集合中。

并查集没有删除的操作，于是换个思路，让p点在其原来的集合中的影响为0，在开辟一个新的节点作为编号为p的节点，加入q所在集合。

用id[i]=ii代表数字i当前的位置，每次删除把id[i]指向其他位置，原来的位置舍去。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int MAXN = 200000+100;int parent[MAXN];int cnt[MAXN];int sum[MAXN];int n, m;int dex;int id[MAXN];void make_set(){dex = n;for (int i = 0; i <= n; i++){parent[i] = i;cnt[i] = 1;sum[i] = i;id[i] = i;}}int find_set(int t){if (parent[t] == t)return t;elsereturn parent[t] = find_set(parent[t]);}void union_set(int a, int b){int t1 = find_set(a);int t2 = find_set(b);if (t1 != t2){parent[t2]=t1;cnt[t1] += cnt[t2];sum[t1] += sum[t2];}}void move(int a){int p = find_set(id[a]);cnt[p]--;sum[p] -= a;         //消除该位置的点在原来集合中的影响id[a] = ++dex;       //开辟新的位置表示这个编号的点parent[dex] = dex;cnt[dex] = 1;sum[dex] = a;}int main(){while (scanf("%d%d", &n, &m) != EOF){make_set();int op;int a, b;while (m--){scanf("%d", &op);if (op == 1){scanf("%d%d", &a, &b);union_set(id[a], id[b]);}else if (op == 2){scanf("%d%d", &a, &b);int t1 = find_set(id[a]);int t2 = find_set(id[b]);if (t1 != t2){move(a);              //从原来集合中移除union_set(id[a], id[b]);   //合并到新的集合中}}else{scanf("%d", &a);int p = find_set(id[a]);printf("%d %d\n", cnt[p], sum[p]);}}}}