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;

}

巴特西

Cube Stacking 来源:洛谷

题目

题目oj(洛谷)

题解

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