Restructuring Company

Even the most successful company can go through a crisis period when you have to make a hard decision — to restructure, discard and merge departments, fire employees and do other unpleasant stuff. Let‘s consider the following model of a company.

There are n people working for the Large Software Company. Each person belongs to some department. Initially, each person works on his own project in his own department (thus, each company initially consists of n

However, harsh times have come to the company and the management had to hire a crisis manager who would rebuild the working process in order to boost efficiency. Let‘s use team(person) to represent a team where person person works. A crisis manager can make decisions of two types:

1. Merge departments team(x) and team(y) into one large department containing all the employees of team(x) and team(y), where x and y (1?≤?x,?y?≤?n) — are numbers of two of some company employees. If team(x) matches team(y), then nothing happens.
2. Merge departments team(x),?team(x?+?1),?...,?team(y), where x
   and y (1?≤?x?≤?y?≤?n) — the numbers of some two employees of the company.

At that the crisis manager can sometimes wonder whether employees x and y (1?≤?x,?y?≤?n) work at the same department.

Help the crisis manager and answer all of his queries.

Input

The first line of the input contains two integers n and q (1?≤?n?≤?200?000, 1?≤?q?≤?500?000) — the number of the employees of the company and the number of queries the crisis manager has.

Next q lines contain the queries of the crisis manager. Each query looks like type x y, where . If type?=?1 or type?=?2, then the query represents the decision of a crisis manager about merging departments of the first and second types respectively. If type?=?3, then your task is to determine whether employees xand y work at the same department. Note that x can be equal to y in the query of any type.

Output

For each question of type 3 print "YES" or "NO" (without the quotes), depending on whether the corresponding people work in the same department.

Example

8 6
3 2 5
1 2 5
3 2 5
2 4 7
2 1 2
3 1 7

NO
YES
YES

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}

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!

ac代碼：

#include<stdio.h>
int f[200005],next[200005];
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;
}
int main()
{
    int n,q,x,y,z,i,j;
    scanf("%d%d",&n,&q);
    for(i=1;i<=n;i++){
        f[i]=i;
        next[i]=i+1;
    }
    for(i=1;i<=q;i++){
        scanf("%d%d%d",&x,&y,&z);
        if(x==1) join(y,z);
        else if(x==2){
            int fz=find(z);
            for(j=y;j<=z;){
                f[find(j)]=fz;
                int t=j;
                j=next[j];
                next[t]=next[z];    //區間合並
            }
        }
        else{
            if(find(y)==find(z)) printf("YES\n");
            else printf("NO\n");
        }
    }
    return 0;
}

#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;    //並查集刪除操作
}
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;
}

Restructuring Company和Almost Union-Find 並查集的區間合並與並查集的刪除