uva11987 - Almost Union-Find 并查集删除元素

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

1是合并p和q的集合，2是把p这个点移动到集合q所在的集合中，3是询问p所在的集合元素个数和元素和。

1，3简单，主要是2。不能直接改变pa[p]，把p接到q的集合中，这样的话若p有子结点，那些子结点就找不到原来的根结点了。为了能让p的子结点找到根结点，pa[p]不能改变。那么可以新建一个结点代表p，给p编号一个新的id，那么p原来的编号的作用就只有让他原来的子结点找到根结点了，以后要修改p的话都使用新的id。把p接到q集合中相当于把p从本来的集合中删掉，建立一个新结点id[p]，在把这个新结点的集合和q的集合合并。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#define INF 0x3f3f3f3f
#define MAXN 200010
#define MAXM 510
#define eps 1e-9
#define pi 4*atan(1.0)
#define pii pair<int,int>
using namespace std;;
int N,M,pa[MAXN],sum[MAXN],num[MAXN],id[MAXN];
int find(int x){
return x==pa[x]?x:pa[x]=find(pa[x]);
}
void Union(int p,int q){
int x=find(p),y=find(q);
if(x!=y){
sum[y]+=sum[x];
num[y]+=num[x];
pa[x]=y;
}
}
void move(int p){
int x=find(id[p]);
sum[x]-=p;
num[x]--;
id[p]=++N;
sum[id[p]]=p;
num[id[p]]=1;
pa[id[p]]=id[p];
}
int main(){
freopen("in.txt","r",stdin);
while(scanf("%d%d",&N,&M)!=EOF){
for(int i=1;i<=N;i++){
id[i]=i;
pa[i]=i;
sum[i]=i;
num[i]=1;
}
while(M--){
int type,p,q;
scanf("%d",&type);
if(type==1){
scanf("%d%d",&p,&q);
Union(id[p],id[q]);
}
if(type==2){
scanf("%d%d",&p,&q);
if(find(id[p])!=find(id[q])){
move(p);
Union(id[p],id[q]);
}
}
if(type==3){
scanf("%d",&p);
int x=find(id[p]);
printf("%d %d\n",num[x],sum[x]);
}
}
}
return 0;
}