POJ 1988 Cube Stacking (并查集）

ACM ICPC 2018 World Finals

Language:

Cube Stacking

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 26668		Accepted: 9350
Case Time Limit: 1000MS

DescriptionFarmer 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.
OutputPrint the output from each of the count operations in the same order as the input file.
Sample Input6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4
Sample Output1
0
2
SourceUSACO 2004 U S Open

很基础的带权并查集， 只要维护好当前的深度， 这一堆一共有多少个方块就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

#define MAX 30030

struct node{
int fa, dep, sum;
}m_arr[MAX];

void init(){
int i;
for(i = 0; i < MAX; i++){
m_arr[i].fa = i;
m_arr[i].dep = 0;
m_arr[i].sum = 1;
}
}

int get(int num){
int tmp = num;
while(m_arr[num].fa != num){
num = m_arr[num].fa;
m_arr[tmp].dep += m_arr[num].dep;
}
m_arr[tmp].fa = num;
return num;
};

void merg(int a, int b){
int fa, fb;
fa = get(a);
fb = get(b);
if(fa != fb){
m_arr[fa].dep = m_arr[fb].sum;
m_arr[fa].fa = fb;
m_arr[fb].sum = m_arr[fb].sum + m_arr[fa].sum;
}
}

int main()
{
int num, i, a, b;
char ord;
scanf("%d", &num);
init();
for(i = 0; i < num; i++){
getchar();
scanf("%c", &ord);
if(ord == 'M'){
scanf("%d%d", &a, &b);
merg(a, b);
}
else{
scanf("%d", &a);
get(a);
printf("%d\n", m_arr[a].dep);
}
}

return 0;
}