POJ 1988 Cube Stacking

题目 - Cube Stacking

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:
moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.

Input

* Line 1: A single integer, P

* Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a ‘M’ for a move operation or a ‘C’ for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.

Output

Print the output from each of the count operations in the same order as the input file.

Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4

Sample Output

1
0
2

题意:

有若干个方块，经p次操作后，在x方块下面的方块有多少个，M操作—>将包含x方块的堆移到含y的堆上，C操作—>输出x方块下方方块的数目。

思路：

• x方块下方方块数目=总数目-上方方块数目

• 如图所示

  

AC代码:

#include<iostream>
using namespace std;
int f[30010],cnt[30010],up[30010];
int find(int x)
{
    int y;
    if(x!=f[x])
    {
       y=f[x];
       f[x]=find(y);
       up[x]+=up[y];
    }
    return f[x];
}
void Union(int x,int y)
{
    int px=find(x);
    int py=find(y);
    if(px==py)
    return ;
    front[py]=px;
    up[py]=cnt[px];
    cnt[px]+=cnt[py];
}
int main()
{
    int p,i,j,x,y;
    char a;
    cin>>p;
    for(i=1;i<=30000;i++)
    {
        f[i]=i;
        cnt[i]=1;
        up[i]=0;
    }
    while(p--)
    {
        cin>>a;
        if(a=='M')
        {
            scanf("%d%d",&x,&y);
            Union(x,y);
        }
        else
        {
            scanf("%d",&x);
            int px=find(x);
            printf("%d\n",cnt[px]-up[x]-1);
        }
    }
    return 0;
}

• 还有一种思路是把父亲放在下面，过程与上图类似，只不过是倒过来了。

AC代码：

#include<iostream>
using namespace std;
int f[30010],cnt[30010],dis[30010];
int find(int x)
{
    int y;
    if(x!=f[x])
    {
        y=f[x];
        f[x]=find(f[x]);
        dis[x]+=dis[y];
    }
    return f[x];
}
void Union(int x,int y)
{
    int px=find(x);
    int py=find(y);
    if(px==py)
    return ;
    f[px]=py;
    dis[px]+=cnt[py];
    cnt[py]+=cnt[px];
}
int main()
{
    int p,i,j,x,y;
    char a;
    cin>>p;
    for(i=1;i<=30000;i++)
    {
        front[i]=i;
        dis[i]=0;
        cnt[i]=1;
    }
    while(p--)
    {
        cin>>a;
        if(a=='M')
        {
            scanf("%d%d",&x,&y);
            Union(x,y);
        }
        else
        {
            scanf("%d",&x);
            find(x);
            cout<<dis[x]<<endl;
        }
    }
    return 0;
}