UVA 11987 - Almost Union-Find

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

Output for the Sample Input

3 12
3 7
2 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}

 

 

本题核心的是并查集的删除，按通常的并查集的做法，集合用某一元素的值表示，若删除此元素，则此元素所在的集合无法寻求根元素。

为解决这个问题，将元素映射为一个id,集合的根用id表示，某一个元素删除后跟存储的是id所以集合任仍在，当某一元素添加到一个集合中是，为该元素分配一个唯一确定的id。以区别在被删除集合中的该元素的id.

 

 

 

 

#include <iostream>
#include "stdio.h"

using namespace std;

const int MAX = 2000000+10;


int r[MAX],id[MAX/2],size[MAX],sum[MAX],cnt;

int find_r(int p)
{
    if(r[p] == p)return p;
    return r[p] = find_r(r[p]);

}

void uni(int p, int q)
{
    int rp = find_r(id[p]);
    int rq = find_r(id[q]);
    if(rp != rq)
    {
        r[rp] = rq;
        size[rq] += size[rp];
        sum[rq] += sum[rp];
    }

}

void move(int p, int q)
{
    int rp = find_r(id[p]);
    int rq = find_r(id[q]);

    --size[rp];
    sum[rp] -= p;

    id[p] = ++cnt;
    r[cnt] = rq;
    sum[rq] += p;
    ++size[rq];

}


int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m) != -1)
    {
        for(int i = 1; i <= n; i++)
        {
            r[i] = i;
            size[i] = 1;
            sum[i] = i;
            id[i] = i;
        }
        cnt = n;



        int symb,p,q;
        while(m--)
        {
            scanf("%d",&symb);
            if(symb == 1)
            {
                scanf("%d%d",&p,&q);
                uni(p,q);
            }
            else if(symb == 2)
            {
                scanf("%d%d",&p,&q);
                move(p,q);
            }
            else if(symb == 3)
            {
                scanf("%d",&p);
                int f = find_r(id[p]);
                printf("%d %d\n",size[f],sum[f]);
            }
        }
    }

    //cout << "Hello world!" << endl;
    return 0;
}