UVA 11987 Almost Union-Find（带有删除操作的并查集）

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

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}

 

 

题意：初始有N个集合，分别为 1 ，2 ，3 .....n。有三种操件
1 ：p q 合并元素p和q的集合
2 ：p q 把p元素移到q集合中
3 ：p 输出p元素集合的个数及全部元素的和。

 

思路： 对于1,3步骤，找两个数组记录即可。

对于2的删除步骤，不能直接把p的根指向q，因为p如果是某个集合的根，那么便会对其他节点产生影响。

所以需要做到p这个节点对集合“无影响”，可以新开一个节点表示p已经从原来的集合中脱离开来，故多开一个数组id[p]表示p节点现在的编号

 

#include
#include
const int maxn = 200010;

int fa[maxn],cnt[maxn],id[maxn];//集合，个数，编号（这个不能缺少）
long long sum[maxn];//总和
int n,m,dep;

void init()
{
    for(int i = 0; i <= n; i ++)
    {
        sum[i] = fa[i] = id[i] = i;
        cnt[i] = 1;
    }
    dep = n;
}

int find(int x)
{
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void unite(int x,int y)
{
    int fx = find(x);
    int fy = find(y);
    fa[fx] = fy;
    sum[fy] += sum[fx];
    cnt[fy] += cnt[fx];
}

void move(int x)
{
    int fx = find(id[x]);     //注意是id[x]而非x，因为一旦x点移走后，它的编号就更新了
    sum[fx] -= x;            //消除x节点的影响
    cnt[fx] --;              //同上
    id[x] = ++ dep;
    sum[id[x]] = x;
    cnt[id[x]] = 1;
    fa[id[x]] = id[x];
}

int main()
{
//  freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int op,x,y;
    while (~scanf("%d%d",&n,&m))
    {
        init();
        while(m--)
        {
            scanf("%d",&op);
            if(op == 1)
            {
                scanf("%d%d",&x,&y);
                if(find(id[x]) != find(id[y]))
                    unite(id[x],id[y]);
            }
            else if(op == 2)
            {
                scanf ("%d%d",&x,&y);
                if(find(id[x])!= find (id[y]))
                    move(x),unite(id[x],id[y]);
            }
            else
            {
                scanf("%d",&x);
                int fx = find(id[x]);
                printf("%d %lld\n",cnt[fx],sum[fx]);
            }
        }
    }
    return 0;
}