Almost Union-Find 并查集的删除

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}

ac代码：

#include<stdio.h>
int f[200005],id[200005],c[200005],sum[200005];
int dex;
int find(int x)
{
    return f[x]==x?x:f[x]=find(f[x]);
}
void join(int x,int y)
{
    int fx=find(x),fy=find(y);
    if(fx!=fy){
        f[fy]=fx;
        c[fx]+=c[fy];
        sum[fx]+=sum[fy];
    }
}
void del(int x)
{
    int fx=find(id[x]);
    c[fx]--;
    sum[fx]-=x;
    id[x]=++dex;
    f[dex]=dex;
    c[dex]=1;
    sum[dex]=x;    //并查集删除操作
}
int main()
{
    int n,q,x,y,z,i;
    while(~scanf("%d%d",&n,&q)){
        dex=n;
        for(i=1;i<=n;i++){
            f[i]=i;
            id[i]=i;
            c[i]=1;
            sum[i]=i;
        }
        for(i=1;i<=q;i++){
            scanf("%d",&x);
            if(x==1){
                scanf("%d%d",&y,&z);
                join(id[y],id[z]);
            }
            else if(x==2){
                scanf("%d%d",&y,&z);
                int fy=find(id[y]);
                int fz=find(id[z]);
                if(fy!=fz){
                    del(y);
                    join(id[y],id[z]);
                }
            }
            else{
                scanf("%d",&y);
                int fy=find(id[y]);
                printf("%d %d\n",c[fy],sum[fy]);
            }
        }
    }
    return 0;
}