标签：acm
题链;http://codeforces.com/problemset/problem/251/B
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:
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.
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.
If the situation that is described in the statement is possible, print "YES" (without the quotes), otherwise print "NO" (without the quotes).
4 1 2 3 4 1 1 2 3 4
NO
4 1 4 3 1 2 3 4 2 1
YES
4 3 4 3 1 2 3 4 2 1
YES
4 2 4 3 1 2 2 1 4 3
YES
4 1 4 3 1 2 2 1 4 3
NO
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：感觉题目完全看不懂啊，，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。并且在这k次变换过程中，p数列不能等于s数列。p数列一开始就为s数列也不行。
做法：因为两个变换是相反的，所以可以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可以达到s数列，然后第二种变换要多少步，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)//一开始就一样
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
*/
版权声明：本文为博主原创文章，未经博主允许不得转载。
cf 251 B Playing with Permutations 暴力 分类讨论
标签：acm
原文地址：http://blog.csdn.net/u013532224/article/details/46794019
