Cube Stacking P0J 1988（加权并查集）

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

Sample Outpu1

0

2

题目意思：有n块方块，p次操作，两种类型，'M a b' 意思是将含有方块a的堆挪到b上：‘C a’意思是查询方块a下面有几个方块。
解题思路：这道题要使用并查集来进行计算。首先为什么会考虑使用并查集呢？因为每一次操作都是把含有a的堆挪到b上，这个时候将会把b包含到这个堆里面，重新组合成一个新的堆，这是什么？状态划分。这道题的难点主要在于虽然在了同一个集合中，但要询问这个集合内的关系，查询a下面几个方块，就可以看成其在集合中的深度。sum[n]数组来统计当前堆n中方块个数。under[n]数组来统计当前n下面的方块个数。在进行计算的时候，路径压缩和合并均更新under的值。未进行路径压缩的时候，方块n所记录的under并非final value，仅仅只是包含了到root[n]的方块个数。通过路径压缩的过程，不断的增加当前n的根节点（递归）的under值，求出最终的under值。进行路径合并的时候，更新sum值和under值。当一个堆放在另一个堆上时，被移动的堆的under值一定为0, 此时更新under值为另一个堆的根的sum值，即计算出了此处的部分under值。然后更新合并堆的sum值。

 #include<cstdio>

 #include<cstring>

 #define maxs 30010

 using namespace std;

 int pre[maxs];

 int sum[maxs];

 int under[maxs];

 void init()

 {

     int i;

     for(i=; i<maxs; i++)

     {

         pre[i]=i;

         under[i]=;

         sum[i]=;

     }

 }

 int Find(int x)

 {

     int t;

     if(x==pre[x])

     {

         return x;

     }

     t=Find(pre[x]);

     under[x]+=under[pre[x]];

     pre[x]=t;

     return pre[x];

 }

 void Union(int x,int y)

 {

     int a=Find(x);

     int b=Find(y);

     if(a==b)

     {

         return ;

     }

     pre[a]=b;

     under[a]=sum[b];///将a堆放在b堆之上,更新此时a堆下面的数量

     sum[b]+=sum[a];///将a堆和b堆合为一个整体，更新整体新堆的数量

 }

 int main()

 {

     int n;

     char q;

     int a,b;

     scanf("%d",&n);

     getchar();

     init();

     while(n--)

     {

         scanf("%c",&q);

         if(q=='M')

         {

             scanf("%d%d",&a,&b);

             getchar();

             Union(a,b);

         }

         else if(q=='C')

         {

             scanf("%d",&a);

             getchar();

             b=Find(a);

             printf("%d\n",under[a]);

         }

     }

     return ;

 }

巴特西

Cube Stacking P0J 1988（加权并查集）

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