UVa11581 - Grid Successors(模拟)

最新推荐文章于 2018-04-14 19:59:36 发布

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最新推荐文章于 2018-04-14 19:59:36 发布

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本文介绍了一个关于3x3矩阵变换的问题，通过定义特定的变换规则，求解矩阵达到自身不变状态所需的变换次数。文章提供了完整的C++代码实现，并详细解释了输入输出样例及解决思路。

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Consider a 3 x 3 grid of numbers g where each cell contains either a 0 or a 1.We define a function f that transforms such a grid: each cell of thegrid f(g) is the sum (modulo 2) of its adjacent cells in g(two cells are considered adjacent if and only if they share a common side).

Furthermore, we define f ⁽ⁱ⁾(g) recursivelyas f ⁽⁰⁾(g) = g andf ⁽ⁱ⁺¹⁾(g) = f(f ⁽ⁱ⁾(g))(where i ≥ 0).Finally, for any grid h, let k_g(h) be the number ofindices i such that h = f ⁽ⁱ⁾(g) (we may have k_g(h) = ∞).Given a grid g, your task is to compute the greatest index isuch that k_g(f ⁽ⁱ⁾(g)) is finite.

Input begins with the number of test cases on its own line.Each case consists of a blank line followed by three lines of three characters, each either 1 or 0.The j'th characterof the i'th row of the test case is the value in the j'th cellof the i'th row of the grid g.

For each test case, output the greatest index i such thatk_g(f ⁽ⁱ⁾(g)) is finite.If there is no such index, output -1.

Sample input

Sample output

3
0
-1

题意：一个3x3的矩阵，求经过多少次变换，原矩阵和新矩阵一样。变换是指将相邻的元素（上、下、左、右）相加的和对2取模。

思路：按照题意，求出新矩阵，将原矩阵与新矩阵相比，是否相等，不相等继续变换，直到相等为止

#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 3;

int a[MAXN][MAXN], b[MAXN][MAXN];

void input()
{
	char buf[MAXN + 1];
	for (int i = 0; i < MAXN; i++) {
		scanf("%s", buf);
		for (int j = 0; j < MAXN; j++) {
			a[i][j] = buf[j] - '0';
		}
	}
}

void transform()
{
	for (int i = 0; i < MAXN; i++) {
		for (int j = 0; j < MAXN; j++) {
			int sum = 0;
			if (j - 1 >= 0) sum += a[i][j - 1];
			if (i - 1 >= 0) sum += a[i - 1][j];
			if (j + 1 < MAXN) sum += a[i][j + 1];
			if (i + 1 < MAXN) sum += a[i + 1][j];
			
			b[i][j] = sum % 2;
		}
	}
}

void solve()
{
	int ans = 0;
	
	while (1) {
		transform();
		if (memcmp(a, b, sizeof(a)) == 0) break;
		ans++;
		memcpy(a, b, sizeof(b));
	}
	
	printf("%d\n", ans - 1);
}

int main(int argc, char **argv)
{
	#ifndef ONLINE_JUDGE
		freopen("e:\\uva_in.txt", "r", stdin);
	#endif
	
	int cas;
	scanf("%d", &cas);
	while (cas--) {
		input();
		solve();
	}

	return 0;
}