poj 1988（并查集）Cube Stacking

Cube Stacking

Time Limit: 2000MS Memory Limit: 30000K

Total Submissions: 24968 Accepted: 8740

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：

有一些碟子，有两种操作，一是把x所在堆的所有碟子都叠到y所在的碟子上，二是询问x所下面有多少个碟子。

思路：

这是一道带权的并查集，跟之前做的差不多。设两个权值cnt[x]代表以x为根结点的堆碟子的个数，dist[x]表示x这碟子到根结点的距离。求x下面有多少碟子就等于cnt[pa[x]] - dist[x] - 1;

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int M = 30005;
int pa[M],cnt[M],dist[M];
void Init()
{
for (int i = 1;i <= M;i ++)
pa[i] = i,cnt[i] = 1,dist[i] = 0;
}

int findset (int x)
{
if (x != pa[x])
{
int root = findset (pa[x]);
dist[x] += dist[pa[x]];
return pa[x] = root;
}
else return x;
}
void Unin(int a,int b)
{
int af = findset (a);
int bf = findset (b);
pa[bf] = af;
dist[bf] = cnt[af];
cnt[af] += cnt[bf];
}
int main ()
{
int m,x,y;
char op[3];
while (~scanf ("%d",&m))
{
Init();
while (m --)
{
scanf ("%s",op);
if (op[0] == 'M')
{
scanf ("%d%d",&x,&y);
Unin(x,y);
}
else
{
scanf ("%d",&x);
int af = findset (x);
printf ("%d\n",cnt[af]-dist[x]-1);
}

}
}
return 0;
}