h: 0px; "> 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).
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
3 7
2 8
真是写残了，脑子都不转了，还好终于写过了，这道题是用到并查集，1,3比较好实现，就是2
不太好实现，要把p加入q就得让p跟原来的树撇清关系，因为p可能是根节点，这样会让p整个集合归入q不合题意，所以需要一个数组这里用newid每次将p move掉然后给p一个新的id重新初始化一切有关的，
还有就是fa数组保存的并不是父节点，要找父节点还是要用getf函数去找
代码：
#include <iostream> using namespace std; long long sum[200005]; int count[200005],fa[200005],newid[200005]; int n,m; void init() { for(int i=1;i<=n;i++) fa[i]=i,count[i]=1,sum[i]=(long long)i,newid[i]=i; } int getf(int x) { if(fa[x]!=x)fa[x]=getf(fa[x]); return fa[x]; } void merge(int x,int y) { int xx=getf(newid[x]),yy=getf(newid[y]); fa[yy]=xx; sum[xx]+=sum[yy]; count[xx]+=count[yy]; } void move(int x) { int xx=getf(newid[x]); count[xx]--; sum[xx]-=(long long)x; newid[x]=++n;// fa[n]=n;// count[n]=1;///注意新指向的id父亲变成自己，成员和是x，这里就要求传入的参数就是输入的p本身，数量是1 sum[n]=(long long)x;///此四行代表给x赋一个新的id以后他就通过这个id来完成各指令 这样原来的x就被切断了 不会保证他的儿子随着x合并到新的群体 } int main() { int op,p,q; while(cin>>n>>m){ init(); while(m--) { cin>>op; if(op==3) { cin>>p; //cout<<newid[p]<<endl; cout<<count[getf(newid[p])]<<' '<<sum[getf(newid[p])]<<endl;///用getf } else { cin>>p>>q; if(getf(newid[p])==getf(newid[q]))continue;///这也是 if(op==1)merge(p,q); else { move(p); merge(q,p); } } } } }
