poj1988_Cube Stacking

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 24130		Accepted: 8468
Case Time Limit: 1000MS

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

唉，一个并查集的题，改了好久。。。终于a了。

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

 using namespace std;

 int up[];//该数上面的数
 int n[];//该集合总数
 int a[];//保存他的父节点

 int father(int t){
     if(t==a[t]){
         return t;
     }
     int fa=a[t];
     a[t]=father(a[t]);
     up[t]+=up[fa];

     return a[t];
 }

 int main()
 {
     int p;
     //char c;
     int t1,t2;
     int t3;
 while(scanf("%d",&p)!=EOF){
     for(int i=;i<=p;i++){
         a[i]=i;
         up[i]=;
         n[i]=;
     }
     char s[];
     for(int i=;i<p;i++){
         scanf("%s",s);
         if(s[]=='M'){
             scanf("%d%d",&t1,&t2);
             int fa=father(t1);
             int fb=father(t2);
             if(fa!=fb){
                 a[fb]=fa;
                 up[fb]+=n[fa];
                 n[fa]+=n[fb];
             }
         }else{
             scanf("%d",&t3);
             int ans=father(t3);
             printf("%d\n",n[ans]-up[t3]-);
         }
     }
 }
     return ;
 }