cf 251 B Playing with Permutations 暴力 分类讨论

题链;http://codeforces.com/problemset/problem/251/B

B. Playing with Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Petya likes permutations a lot. Recently his mom has presented him permutation q1, q2, ..., qn of
length n.

A permutation a of length n is
a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ n),
all integers there are distinct.

There is only one thing Petya likes more than permutations: playing with little Masha. As it turns out, Masha also has a permutation of length n.
Petya decided to get the same permutation, whatever the cost may be. For that, he devised a game with the following rules:

• Before the beginning of the game Petya writes permutation 1, 2, ..., n on the blackboard. After that Petya makes exactly k moves,
  which are described below.
• During a move Petya tosses a coin. If the coin shows heads, he performs point 1, if the coin shows tails, he performs point 2.
  1. Let's assume that the board contains permutation p1, p2, ..., pn at
     the given moment. Then Petya removes the written permutation p from the board and writes another one instead: pq1, pq2, ..., pqn.
     In other words, Petya applies permutation q(which he has got from his mother) to permutation p.
  2. All actions are similar to point 1, except that Petya writes permutation t on the board, such that: tqi = pi for
     all i from 1 to n.
     In other words, Petya applies a permutation that is inverse to q to permutation p.

We know that after the k-th move the board contained Masha's permutation s1, s2, ..., sn.
Besides, we know that throughout the game process Masha's permutation never occurred on the board before the k-th
move. Note that the game has exactly k moves, that is, throughout the game the coin was tossed exactly k times.

Your task is to determine whether the described situation is possible or else state that Petya was mistaken somewhere. See samples and notes to them for a better understanding.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 100).
The second line contains n space-separated integers q1, q2, ..., qn(1 ≤ qi ≤ n)
— the permutation that Petya's got as a present. The third line contains Masha's permutation s, in the similar format.

It is guaranteed that the given sequences q and s are
correct permutations.

Output

If the situation that is described in the statement is possible, print "YES" (without the quotes), otherwise print "NO"
(without the quotes).

Sample test(s)

input

4 1
2 3 4 1
1 2 3 4

output

NO

input

4 1
4 3 1 2
3 4 2 1

output

YES

input

4 3
4 3 1 2
3 4 2 1

output

YES

input

4 2
4 3 1 2
2 1 4 3

output

YES

input

4 1
4 3 1 2
2 1 4 3

output

NO

Note

In the first sample Masha's permutation coincides with the permutation that was written on the board before the beginning of the game. Consequently, that violates the condition that Masha's permutation never occurred on the board before k moves
were performed.

In the second sample the described situation is possible, in case if after we toss a coin, we get tails.

In the third sample the possible coin tossing sequence is: heads-tails-tails.

In the fourth sample the possible coin tossing sequence is: heads-heads.

题意：感觉题目全然看不懂啊。。突然就冒出了个t。

题意是，给q数列，和s数列。然后p数列初始为1-n。然后通过p[q[i]]=p[i]。或者p[i]=p[q[i]]这两种变换，问有没有可能在k次变换后刚刚p数列为s数列。而且在这k次变换过程中。p数列不能等于s数列。p数列一開始就为s数列也不行。

做法：由于两个变换是相反的。所以能够通过两次分别两种变换来抵消。计算出p通过第一种变换要多少步能够达到s数列，然后另外一种变换要多少步。然后分类讨论。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define INF 999999999
#define eps 0.00001
#define LL __int64
#define pi acos(-1.0)

int pp[1010],p[1010];
int q[1010],s[1010];
int main()
{
	int n,k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		for(int i=1;i<=n;i++)
			scanf("%d",&q[i]);
		for(int i=1;i<=n;i++)
			scanf("%d",&s[i]);
		for(int i=1;i<=n;i++)
			pp[i]=p[i]=i;
		int flag;
		int T1=0;
		while(T1<=k)//tou
		{
			flag=1;
			for(int i=1;i<=n;i++)
			{
				if(pp[i]!=s[i])
					flag=0;
			}
			if(flag)
				break;
			for(int i=1;i<=n;i++)
				pp[i]=p[q[i]];
			for(int i=1;i<=n;i++)
				p[i]=pp[i];
			T1++;
		}
		for(int i=1;i<=n;i++)
			pp[i]=p[i]=i;
		int T2=0;
		while(T2<=k)//tou
		{
			flag=1;
			for(int i=1;i<=n;i++)
			{
				if(pp[i]!=s[i])
					flag=0;
			}
			if(flag)
				break;
			for(int i=1;i<=n;i++)
				pp[q[i]]=p[i];
			for(int i=1;i<=n;i++)
				p[i]=pp[i];
			T2++;
		}

		if(T1==0||T2==0)//一開始就一样
			printf("NO\n");
		else if(k==1&&(T1==1||T2==1))//一步一样
			printf("YES\n");
		else if(k!=1&&T1==1&&T2==1)
			printf("NO\n");
		else if(T1==k+1&&T2==k+1)
			printf("NO\n");
		else if((T1-k)%2==0&&T1<=k)
			printf("YES\n");
		else if((T2-k)%2==0&&T2<=k)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}
/*
4 1
2 3 4 1
1 2 3 4

*/