UVA - 11987 Almost Union-Find[并查集 删除]

UVA - 11987

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:

1pq	Union the sets containing p and q. If p and q are already in the same set, ignore this command.
2pq	Move p to the set containing q. If p and q are already in the same set, ignore this command.
3p	Return the number of elements and the sum of elements in the set contain- ing 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).

Output

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

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}

Sample Input

5 7 

1 1 2

2 3 4 

1 3 5 

3 4 

2 4 1 

3 4 

3 3 

Sample Output

3 12 37 28 

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 白书

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删除操作

用一个id，删除时给被删除的点重新分配id，旧的父亲和新的父亲分别更新行了

注意本题权值就是自己

沙茶的查询忘加id了.............

//
//  main.cpp
//  uva11987
//
//  Created by Candy on 10/12/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
const int N=1e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,m,op,p,q;
int fa[N],sum[N],id[N],cnt[N],num=0;
inline int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
inline void unn(int x,int y){
    int f1=find(id[x]),f2=find(id[y]);
    if(f1!=f2){
        fa[f1]=f2;
        sum[f2]+=sum[f1];
        cnt[f2]+=cnt[f1];
    }
}
inline void move(int x,int y){
    int f1=find(id[x]),f2=find(id[y]);
    if(f1!=f2){
        sum[f1]-=x;
        cnt[f1]--;
        id[x]=++num;
        sum[f2]+=x;
        cnt[f2]++;
        fa[id[x]]=f2;
    }
}
int main(int argc, const char * argv[]) {
    while(scanf("%d%d",&n,&m)!=EOF){
        num=n;
        for(int i=1;i<=n;i++) {fa[i]=sum[i]=id[i]=i;cnt[i]=1;}
        for(int i=1;i<=m;i++){
            op=read();
            if(op==1){
                p=read();q=read();
                unn(p,q);
            }else if(op==2){
                p=read();q=read();
                move(p,q);
            }else{
                p=read();
                p=find(id[p]);
                printf("%d %d\n",cnt[p],sum[p]);
            }
        }
    }
    
    return 0;
}