# POJ1988 Cube Stacking（带权并查集）

## Cube Stacking

• Time Limit: 2000MS
• Memory Limit: 30000K
• Total Submissions: 26967
• Accepted: 9443
• 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

### 链接

http://poj.org/problem?id=1988

### 题意

1. $M \ X \ Y$，将$X$方块所在的堆移动至$Y$方块所在的堆的顶部
2. $C \ X$，查询$X$方块下面方块的数量

### 题解

• 合并两个方块$x$和$y$时，总是把下方的方块作为父节点，让$F[x]=y$，每堆的根节点为该堆最下面的一个方块
• 合并$x$和$y$方块所在的两个堆$X$和$Y$时，首先找到$X$堆的根节点$fx(=find(x))$，然后找到$Y$堆的根节点$fy(=find(y))$，让$F[fx]=fy$，$cnt[fx] = sum[fy]$，$sum[fy] += sum[fx]$
• 在查询根节点的路径压缩操作中，对$cnt[x]$进行更新，$cnt[x] += cnt[F[x]]$

### 代码

• Status：Accepted
• Time：922ms
• Memory：1080kB
• Length：982
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 30010;
int F[maxn];
int sum[maxn];
int cnt[maxn];

void init()
{
for (int i = 0; i < maxn; ++i)
{
F[i] = i;
sum[i] = 1;
cnt[i] = 0;
}
}

int find(int root)
{
if (root == F[root])
return root;
int tmp = find(F[root]); // 找到root的根节点
cnt[root] += cnt[F[root]]; // 更新root的方块数
F[root] = tmp; // 更新root的根节点
return tmp;
}

void merge(int u, int v)
{
int fx = find(u), fy = find(v);
if (fx != fy)
{
F[fx] = fy; // x堆放到y堆的上方，y堆最下方元素作为根节点
cnt[fx] = sum[fy]; // 方块fx下面的方块树即为y堆的数量
sum[fy] += sum[fx]; // y堆的方块数更新为两个堆的方块总数
}
}

int main()
{
init();
int n, x, y;
char ch;
scanf("%d", &n);
for (int i = 0; i < n; ++i)
{
cin >> ch;
if (ch == 'M')
{
scanf("%d%d", &x, &y);
merge(x, y);
}
else
{
scanf("%d", &x);
find(x); // 注意更新！！！
printf("%d\n", cnt[x]);
}
}
return 0;
}

The end.
2018年04月15日 星期天

1. @(不高兴) 可能下次遇到这种题还是不会做。