题目

​题目oj(洛谷)​

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

给定N个方块，排成一行，将它们编号1到N。再给出P个操作：

①M i j表示将i所在的那一堆移到j所在那一堆的顶上。

②C i表示一个询问，询问i下面有多少个方块。

•你需要写一个程序来完成这些操作。

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原文链接：https://blog.csdn.net/qq_42434171/article/details/81711250

题解

关于这一道题目,想到使用并查集,真的是大神!!!

在这道题目中,通过阅读英文题目,发现它有并查集的影子!

在移动的时候,它是把A所在的一摞(所有的)挪动到B所在的一摞(所有的)的最上面

这就需要快速地判断哪些是在一摞上.

所以猜想使用并查集;

再想:应该把最底下的哪一个叫做根,因为比较稳定

事实上:并查集并不是仅仅用来判断是不是在一个集合里边,还可以有拓展的用途.

(比如记录某一个点到根节点的距离(在这里,我尚且把箱子的个数抽象为距离))

这个距离可以使用一次一次找爸爸的过程来进行求解, 同时还可以路径压缩

代码

#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
using namespace std;
#define MAX 30009
int fa[MAX];
int dis[MAX];
int num[MAX];
int find(int x)//注意找的时候要顺便进行路径压缩,在这个过程中,dis的值也要进行更改
//由于这样,所以就不许要在单独设计一个找下面有多少个箱子的函数,直接先来一个find,直接输出dis[i]
{
  if (x != fa[x])//自己不是根节点
    //整个return递归采用了回溯的方法
  {
    int ret = find(fa[x]);
    dis[x] += dis[fa[x]];
    fa[x] = ret;
    return ret;
  }
  else
  {
    return x;
  }
}
void move(int x, int y)
{
  int fx = find(x);
  int fy = find(y);
  fa[fx] = fy;
  dis[fx] += num[fy];
  num[fy] += num[fx];
  
}

int main()
{
  int T;
  scanf("%d", &T);
  for (int i = 1; i <= 30000; i++)
  {
    num[i] = 1;
    fa[i] = i;
  }
  while (T--)
  {
    char buf[12];
    int t1, t2;
    scanf("%s", buf);
    if (*buf == 'M')//移动
    {
      scanf("%d%d", &t1, &t2);
      move(t1, t2);
    }
    else//查询
    {
      scanf("%d", &t1);
      find(t1);
      printf("%d\n", dis[t1]);
    }
  }

  return 0;

}