【POJ 1988】 Cube Stacking （带权并查集）

Cube Stacking

Description

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

Source

USACO 2004 U S Open

 

 

【题意】

　　约翰和贝西在玩叠积木，每块积木是大小一样的立方体，将几块积木叠在一起可以得到更高的长
方体。积木是有磁性的，叠起来的积木分不开。
约翰负责把积木叠高，而贝西负责清点积木的高度。一开始共有 N 块积木，编号为 1 到 N，所
有积木都是分开的，没有积木叠在一起。
每次移动积木的时候，约翰会先选择一块积木 X，再选择另一块积木 Y，把 X 搬到 Y 的上方。
如果 X 已经和其它积木叠在一起了，那么应将这叠积木整体移动到 Y 的上方；如果 Y 已经和其它积
木叠在一起了的，假设在 Y 上方最高处的积木为 Z，那么应将 X 所在的那叠积木移动到 Z 的上方。
在约翰辛苦劳动的同时，贝西在一边计算积木已经叠得多高了。约翰和贝西是交替工作的，可惜
贝西不太擅长于数数，请你帮她计算一下，在一些时刻，指定的积木的下方还有多少其他积木。

 

【分析】

　　我这个并查集的权值是他跟父亲的距离，最后做一遍find就知道他跟最低点的距离（即下面有多少个东西），然后我还开多了一个并查集，表示这个东西最上面是什么东西。

　　差不多这样把。

 

代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<queue>
 #include<cmath>
 using namespace std;
 #define Maxn 300010

 int fa[Maxn],d[Maxn],g[Maxn];
 char s[];

 int ffind(int x)
 {
     int y=fa[x];
     if(fa[x]!=x) fa[x]=ffind(fa[x]);
     d[x]+=d[y];
     return fa[x];
 }

 int find_g(int x)
 {
     if(g[x]!=x) g[x]=find_g(g[x]);
     return g[x];
 }

 int main()
 {
     int q;
     scanf("%d",&q);
     for(int i=;i<=;i++) fa[i]=i,g[i]=i,d[i]=;
     for(int i=;i<=q;i++)
     {
         scanf("%s",s);
         if(s[]=='M')
         {
             int x,y;
             scanf("%d%d",&x,&y);
             int z=ffind(x);y=find_g(y);
             fa[z]=y;d[z]=;
             g[find_g(y)]=find_g(x);
         }
         else
         {
             int x;
             scanf("%d",&x);
             ffind(x);
             printf("%d\n",d[x]);
         }
     }
     return ;
 }

[POJ 1988]

2016-10-27 20:25:21