ZOJ 3789 Gears

并查集,

删除节点操作,可以用新建节点代替

维护每个点到跟节点的距离

Gears

Bob has M (1 ≤ N ≤ 4*10⁵) operations to his gears group:

• "L u v". Link gear u and gear v. If two gears link together, the gear groups which the two gears come from will also link together, and it makes a new gear group. The two gears will have different rotations. BTW, Bob won‘t link two
  gears with the same rotations together, such as linking (a, b), (b, c), and (a, c). Because it would shutdown the rotation of his gears group, he won‘t do that.
• "D u". Disconnect the gear u. Bob may feel something wrong about the gear. He will put the gear away, and the gear would become a new gear group itself and may be used again later. And the rest of gears in this group won‘t be separated apart.
• "Q u v". Query the rotations between two gears u and v. It could be "Different", the "Same" or "Unknown".
• "S u". Query the size of the gears, Bob wants to know how many gears there are in the gear group containing the gear u.

Since there are so many gears, Bob needs your help.

Input

Input will consist of multiple test cases. In each case, the first line consists of two integers N and M. Following M lines, each line consists of one
of the operations which are described above. Please process to the end of input.

Output

For each query operation, you should output a line consist of the result.

Sample Input

3 7
L 1 2
L 2 3
Q 1 3
Q 2 3
D 2
Q 1 3
Q 2 3
5 10
L 1 2
L 2 3
L 4 5
Q 1 2
Q 1 3
Q 1 4
S 1
D 2
Q 2 3
S 1

Sample Output

Same
Different
Same
Unknown
Different
Same
Unknown
3
Unknown
2

Hint

Link (1, 2), (2, 3), (4, 5), gear 1 and gear 2 have different rotations, and gear 2 and gear 3 have different rotations, so we can know gear 1 and gear 3 have the same rotation, and we
didn‘t link group (1, 2, 3) and group (4, 5), we don‘t know the situation about gear 1 and gear 4. Gear 1 is in the group (1, 2, 3), which has 3 gears. After putting gear 2 away, it may have a new rotation, and the group becomes (1, 3).

Source: ZOJ Monthly, June 2014

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=600600;

int Sz[maxn],dist[maxn],fa[maxn],now[maxn],next;
int n,m;

void init(int n,int m)
{
    memset(dist,0,sizeof(dist));
    next=n+1;
    for(int i=1;i<=n+m;i++)
    {
        Sz[i]=1;fa[i]=i;
        if(i<=n) now[i]=i;
    }
}

int Find(int x)
{
    if(fa[x]==x) return x;
    int temp=Find(fa[x]);
    dist[x]+=dist[fa[x]];///到gen节点的距离
    return fa[x]=temp;
}

void Union(int x,int y)
{
    int X=Find(x),Y=Find(y);
    if(X==Y) return ;
    fa[X]=Y;
    Sz[Y]+=Sz[X];
    dist[X]=dist[Y]+1;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init(n,m);
        char op[5];int a,b;
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='L')
            {
                scanf("%d%d",&a,&b);
                int na=now[a],nb=now[b];
                Union(na,nb);
            }
            else if(op[0]=='D')
            {
                scanf("%d",&a);
                int na=now[a];
                int A=Find(na);
                Sz[A]--;Sz[na]=0;
                now[a]=next++;
            }
            else if(op[0]=='S')
            {
                scanf("%d",&a);
                int na=now[a];
                int A=Find(na);
                printf("%d\n",Sz[A]);
            }
            else if(op[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                int na=now[a],nb=now[b];
                int A=Find(na),B=Find(nb);
                if(A!=B) puts("Unknown");
                else
                {
                    if ((dist[na]-dist[nb])&1) puts("Different");
                    else puts("Same");
                }
            }
        }
    }
    return 0;
}

ZOJ 3789 Gears