pSort

pSort

题面翻译

题目描述

给定一个长度为 $n$ 的数列 { a n } \{a_n\} {an​}，初始时 { a n } = { 1 , 2 , … , n } \{a_n\} = \{1, 2, \dots, n\} {an​}={1,2,…,n}。位置 $i$ 上的数可以和位置 $i\pm d_i$ 上的数交换。给定一个 $1\sim n$ 的全排列，问初始的数列可否交换成给定的形式。

输入格式

第一行一个整数 $n$。

第二行 $n$ 个互不相同的整数表示目标数列。

第三行 $n$ 个整数表示 $d_1,d_2,\dots ,d_n$。

输出格式

如果能交换到给定样式，输出 YES，否则输出 NO。

数据范围

$1\le n\le 100$

题目描述

One day $ n $ cells of some array decided to play the following game. Initially each cell contains a number which is equal to it’s ordinal number (starting from $ 1 $ ). Also each cell determined it’s favourite number. On it’s move $ i $ -th cell can exchange it’s value with the value of some other $ j $ -th cell, if $ |i-j|=d_{i} $ , where $ d_{i} $ is a favourite number of $ i $ -th cell. Cells make moves in any order, the number of moves is unlimited.

The favourite number of each cell will be given to you. You will also be given a permutation of numbers from $ 1 $ to $ n $ . You are to determine whether the game could move to this state.

输入格式

The first line contains positive integer $ n $ ( $ 1<=n<=100 $ ) — the number of cells in the array. The second line contains $ n $ distinct integers from $ 1 $ to $ n $ — permutation. The last line contains $ n $ integers from $ 1 $ to $ n $ — favourite numbers of the cells.

输出格式

If the given state is reachable in the described game, output YES, otherwise NO.

样例 #1

样例输入 #1

5
5 4 3 2 1
1 1 1 1 1

样例输出 #1

YES

样例 #2

样例输入 #2

7
4 3 5 1 2 7 6
4 6 6 1 6 6 1

样例输出 #2

NO

样例 #3

样例输入 #3

7
4 2 5 1 3 7 6
4 6 6 1 6 6 1

样例输出 #3

YES

思路：对于每一个数，想要到达它原来的位置的话，我们应该能够确保它们在一个连通块内，只有这样才能够到达目标位置，因此我们可以想到利用并查集，最后判断i与fa[a[i]]是否一样即可

AC代码：

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int>PII;
const int MOD = 998244353;
const int N = 4e5 + 10;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};


int fa[110];
int a[110];
int d[110];
int finds(int x)
{
	if(x != fa[x]) fa[x] = finds(fa[x]);
	return fa[x];
}

void un(int x, int y)
{
	int o1 = finds(x), o2 = finds(y);
	if(o2 != o1) 
	{
		fa[o2] = fa[o1];
	}
}
int main()
{
	int n;
		cin >> n;
		for(int i = 1; i <= n; i ++) fa[i] = i;
		for(int i = 1; i <= n; i ++) cin >> a[i];
		for(int i = 1; i <= n; i ++) cin >> d[i];
		for(int i = 1; i <= n; i ++)
		{
			if(i > d[i]) un(i, i - d[i]);
			if(i + d[i] <= n) un(i, i + d[i]);
		}
		int f = 0;
		for(int i = 1; i <= n; i ++)
		{
			if(finds(i) != finds(a[i]))
			{
				f = 1;
				break;
			}
		}
		if(f) puts("NO");
		else puts("YES");
}