[思维]奇数码问题

奇数码问题

给定奇数版, 和两种状态A B

问能否有解使板A 变为板 B

例:

1 2 3
0 4 6
7 5 8

1 2 3
4 5 6
7 8 0

---

解: 将板A, B化为一维向量, 

转化为逆序对问题

首先去掉位0

有 1.空格的左右移动不影响整个串的顺序

 2.空格的上下移动必有等价与 swap(s[i], s[i - n - 1]) / swap(s[i], s[i + n - 1]) 

因为n - 1 为偶数, 所以逆序对的改变只能为偶数

由以上推导, 若A, B向量的 逆序对奇 偶性相同, 则必定可以转换得解

/*
	Zeolim - An AC a day keeps the bug away
*/

//pragma GCC optimize(2)
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <sstream>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
typedef long long ll;

const int MAXN = 1e6 + 10;

int arr[MAXN];

int brr[MAXN];

int revnum = 0;

void mergesort(int *arr, int fst, int lst, int *brr)
{
	if(lst - fst > 1)
	{
		int mid = fst + (lst - fst) / 2;
		int p = fst, q = mid, i = fst;
		
		mergesort(arr, fst, mid, brr);
		mergesort(arr, mid, lst, brr);
		
		while(p < mid || q < lst)
		{
			if(q >= lst || (p < mid && arr[p] <= arr[q]))
			{
				brr[i++] = arr[p++];
			}
			else
			{
				brr[i++] = arr[q++];
				revnum += mid - p;
			}
		}
		for(i = fst; i < lst; i++)
				arr[i] = brr[i];
	}
}

int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);     cout.tie(0);
    //freopen("D://test.in", "r", stdin);
    //freopen("D://test.out", "w", stdout);
    
    int n;

    while(cin >> n)
    {
    	for(int i = 0, j = 0, y; i < n * n; ++i)
	    {
	    	cin >> y;

	    	if(y)
	    		arr[j++] = y;
	    }

	    revnum = 0;

	    mergesort(arr, 0, n * n - 1, brr);

	    int ra = revnum;

	    for(int i = 0, j = 0, y; i < n * n; ++i)
	    {
	    	cin >> y;

	    	if(y)
	    		arr[j++] = y;
	    }

	    revnum = 0;

	    mergesort(arr, 0, n * n - 1, brr);

	    int rb = revnum;

	    if((ra & 1) && (rb & 1) || (!(ra & 1) && !(rb & 1)))
	    	cout<<"TAK
";
	    else
	    	cout<<"NIE
";


    }

    



    return 0;
}