uva 11987 并查集

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

这是一道很有的意思的并查集的变式题，其实题目并没有什么特别大的难度，关键就是并查集并没有删除这种操作，而该题正是要将并查集加上删除的操作，菜鸟今天从大牛那里学到了一招就是不需要删除 点，只需要在新建一个结点，。代码被删除的结点现在的位置，保证了删除的结点对原集合的影响降到了0.

题意：

给定一些操作：
1 a b是合并两个数a b(直接用merge(a,b))2 a b是将a从原来所在集合中删除，将a移动到b所在的集合中(delete(a),merge(a,b))3 a 是查询a所在的集合有 多少个数，以及这些数的和，需要打印下面是代码：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int maxn=200050;

int id[maxn],num[maxn];//id是关键记录下现在的结点的位置
int pre[maxn];
char cmd;
long long sum[maxn];
int cnt;
int n,m;

void init(){
for(int i=1;i<=n;i++){
id[i]=pre[i]=sum[i]=i;
num[i]=1;
cnt=n;
}
}

int find(int v){
return pre[v]==v?v:find(pre[v]);
}

void join(int a,int b){
int fx=find(id[a]),fy=find(id[b]);
if(fx!=fy){
pre[fx]=fy;
num[fy]+=num[fx];
sum[fy]+=sum[fx];
}
}

void Delete(int a){
int root=find(id[a]);
num[root]=num[root]-1;
sum[root]-=a;
id[a]=++cnt;
pre[id[a]]=id[a];//最开始脑残忘了写这一步了
num[id[a]]=1;
sum[id[a]]=a;
}

int main(){
while(scanf("%d%d",&n,&m)==2){
int p,q;
init();
for(int i=0;i<m;i++){//分情况讨论
cin>>cmd;
if(cmd=='1')
{
cin>>p>>q;
if(find(id[p])!=find(id[q]))
join(p,q);
}
if(cmd=='2'){
cin>>p>>q;
if(find(id[p])!=find(id[q])){
Delete(p);
join(p,q);
}
}
if(cmd=='3'){
cin>>p;
printf("%d %lld\n", num[find(id[p])], sum[find(id[p])]);
}
}
}
return 0;
}