CH0503 奇数码问题

题意

给出两个$n\ast n$的矩阵，一个起始矩阵，一个目标矩阵，包含$[1,n\ast n-1]$的整数和一个空格，可以上下左右移动该空格。问是否可以通过若干操作，使得起始矩阵变为目标矩阵。

思路

将矩阵的数字从上到下从左到右依次排列为长度为$n\ast n-1$的序列（去掉空格），若初始矩阵和目标矩阵的逆序对数的奇偶性相同，则可以，否则不行。

代码

#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cstdio>
#include <iostream>
using namespace std;

#define fo(i, x, y) for (int i = x; i <= y; ++i)
#define fd(i, x, y) for (int i = x; i >= y; --i)

typedef long long ll;

const int maxnn = 250000 + 5;

int n;
int a[maxnn], b[maxnn];

int getint()
{
	char ch;
	int res = 0, p;
	while (!isdigit(ch = getchar()) && (ch ^ '-'));
	p = ch == '-'? ch = getchar(), -1 : 1;
	while (isdigit(ch))
		res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
	return res * p;
}

ll merge(int l, int r)
{
	if (l >= r) return 0;
	int mid = (l + r) >> 1;
	ll ans = merge(l, mid) + merge(mid + 1, r);
	int pl = l, pr = mid + 1;
	fo(i, l, r)
		if (pr > r || pl <= mid && a[pl] <= a[pr]) b[i] = a[pl++];
			else b[i] = a[pr++], ans += mid - pl + 1;
	fo(i, l, r) a[i] = b[i];
	return ans;
}

int main()
{
	while (~scanf("%d", &n) && n)
	{
		int p;
		fo(i, 0, n - 1)
			fo(j, 0, n - 1)
			{
				a[i * n + j] = getint();
				if (!a[i * n + j]) p = i * n + j;
			}
		fo(i, p, n * n - 2) a[i] = a[i + 1];
		ll sa = merge(0, n * n - 2);
		fo(i, 0, n - 1)
			fo(j, 0, n - 1)
			{
				a[i * n + j] = getint();
				if (!a[i * n + j]) p = i * n + j;
			}
		fo(i, p, n * n - 2) a[i] = a[i + 1];
		ll sb = merge(0, n * n - 2);
		if ((sa & 1) && (sb & 1) || !(sa & 1) && !(sb & 1)) printf("TAK\n");
			else printf("NIE\n");
	}
	return 0;
}