UVA - 11987 - Almost Union-Find （又是并查集~）

UVA - 11987

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!

Source

Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 3. Data Structures :: Fundamental Data Structures :: Exercises: Intermediate
Root :: Prominent Problemsetters :: Rujia Liu
Root :: Rujia Liu's Presents :: Present 3: A Data Structure Contest

思路：1,3比较好实现，2则有点麻烦，2属于并查集的删除操作，需要另设一组real[]数组来确定元素的实际地址，每删除一个元素，就把这个元素放在最后面。。我本来以为直接把p指向q的根就行了，但发现这是错的。如果p是叶子结点，那可以，但如果是某个集合的根呢。我们只是要把这一个元素移掉，如果直接把p指向q的根，它的叶子节点们也过去啦。。。而如果另设一组real数组的话就可以将影响降为0啦。。

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#define LL long longusing namespace std;const int maxn = 200005;int pa[maxn], real[maxn], cnt[maxn];int n, m, vnum;LL sum[maxn];int find(int x) {return pa[x] != x ? pa[x] = find(pa[x]) : x;}void Union(int a, int b) {int a1 = find(real[a]), b1 = find(real[b]);pa[a1] = b1;sum[b1] += sum[a1];cnt[b1] += cnt[a1];}void Move(int a) {int t = find(real[a]);sum[t] -= a, cnt[t]--;real[a] = ++vnum;//并查集的这里到n了，所以要先++，之前后++的样例没过，检查半天=_=||sum[real[a]] = a, cnt[real[a]] = 1, pa[real[a]] = real[a];}int main() {while(scanf("%d %d", &n, &m) != EOF) {for(int i = 0; i <= n; i++) {pa[i] = real[i] = sum[i] = i;cnt[i] = 1;}vnum = n;int ord, p, q;while(m--) {scanf("%d", &ord);if(ord == 1) {scanf("%d %d", &p, &q);if(find(real[p]) != find(real[q])) Union(p, q);} else if(ord == 2) {scanf("%d %d", &p, &q);if(find(real[p]) != find(real[q])) Move(p), Union(p, q);} else if(ord == 3) {scanf("%d", &p);int tmp = find(real[p]);printf("%d %lld\n", cnt[tmp], sum[tmp]);}}}return 0;}