【POJ 1988 Cube Stacking】+ 并查集

Cube Stacking
Time Limit: 2000MS

Memory Limit: 30000K
Total Submissions: 25082

Accepted: 8781
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

AC代码：

#include<cstdio>#include<iostream>using namespace std;int F[30010];int R[30010];int num[30010];int Found(int x){    if(F[x]==x) return x;    int t=F[x];    F[x]=Found(F[x]);    R[x]=R[t]+R[x];    return F[x];}int main(){int q;     while(~scanf("%d",&q)){        for(int i=1;i<=30000;i++) F[i]=i,R[i]=0,num[i]=1;        char s[10];        while(q--){            scanf("%s",s);            if(s[0]=='C'){                int a; scanf("%d",&a);Found(a);                printf("%d\n",R[a]);            }            else{                int a,b; scanf("%d %d",&a,&b);                int na = Found(a),nb = Found(b);                if(na!=nb){                    F[na]=nb;                    R[na] = num[nb];                    num[nb]+=num[na];                }            }        }    }    return 0;}