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.
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.
* 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 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。
1 #include<cstdio> 2 #include<cstring> 3 #define maxs 30010 4 using namespace std; 5 int pre[maxs]; 6 int sum[maxs]; 7 int under[maxs]; 8 void init() 9 { 10 int i; 11 for(i=0; i<maxs; i++) 12 { 13 pre[i]=i; 14 under[i]=0; 15 sum[i]=1; 16 } 17 } 18 int Find(int x) 19 { 20 int t; 21 if(x==pre[x]) 22 { 23 return x; 24 } 25 t=Find(pre[x]); 26 under[x]+=under[pre[x]]; 27 pre[x]=t; 28 return pre[x]; 29 } 30 void Union(int x,int y) 31 { 32 int a=Find(x); 33 int b=Find(y); 34 if(a==b) 35 { 36 return ; 37 } 38 pre[a]=b; 39 under[a]=sum[b];///将a堆放在b堆之上,更新此时a堆下面的数量 40 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 41 } 42 43 int main() 44 { 45 int n; 46 char q; 47 int a,b; 48 scanf("%d",&n); 49 getchar(); 50 init(); 51 while(n--) 52 { 53 scanf("%c",&q); 54 if(q=='M') 55 { 56 scanf("%d%d",&a,&b); 57 getchar(); 58 Union(a,b); 59 } 60 else if(q=='C') 61 { 62 scanf("%d",&a); 63 getchar(); 64 b=Find(a); 65 printf("%d ",under[a]); 66 } 67 } 68 return 0; 69 }