POJ 1988 Cube Stacking并查集

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000K

Total Submissions: 22236 Accepted: 7803

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

题意：初始时有n个堆栈，每个堆栈上面1个cube。(编号都为1 ~ n)，现在定义下面两个操作：

操作’M’ X Y: 按数据结构的栈来操作，将含有X元素的栈的元素，放到元素Y的上面

操作’C’ X：统计第X个cube下面含有多少个cube

思路：分别用under[i]保存第i个cube下面的cube数目；num[x]表示根为x的集合元素数；fat[x]表示元素x所在集合的根。





例如：M X Y的操作变化如下

#include <stdio.h>
#include <fstream>
#include <string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 30005;
int fat[maxn],P,x,y,num[maxn],under[maxn];
char m[10];
int find(int x){
int tmp=fat[x];
if(x!=fat[x])
fat[x]=find(fat[x]),under[x]+=under[tmp];//为什么这里要加上under[tmp],见图一
return fat[x];
}
void Union(int x,int y){
int fx=find(x),fy=find(y);
if(fx==fy) return;
fat[fx]=fy;
under[fx]+=num[fy];
num[fy]+=num[fx];
}
int main(){
freopen("i.txt","r",stdin);
cin>>P;
for(int i=0;i<maxn;i++) fat[i]=i,under[i]=0,num[i]=1;
while(P--){
scanf("%s",m);
if(m[0]=='M'){
scanf("%d%d",&x,&y);
Union(x,y);
}
else{
scanf("%d",&x);
find(x);
cout<<under[x]<<endl;
}
}
return 0;
}