UVA - 1156 Pixel Shuffle （置换+模拟）

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UVA - 1156 Pixel Shuffle （置换+模拟）

2024-07-20 13:56:18 220人阅读

Description

$\epsfbox{p3510a.eps}$

Shuffling the pixels in a bitmap image sometimes yields random looking images. However, by repeating the shuffling enough times, one finally recovers the original images. This should be no surprise, since ``shuffling" means applying a one-to-one mapping (or permutation) over the cells of the image, which come in finite number.

Problem

Your program should read a number n , and a series of elementary transformations that define a ``shuffling" $\phi$ ofnxn images. Then, your program should compute the minimal numberm(m > 0) , such thatm applications of $\phi$ always yield the originalnxn image.

For instance if $\phi$ is counter-clockwise 90^o rotation then m = 4 .

$\epsfbox{p3510b.eps}$

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

Input is made of two lines, the first line is number n (2 $\le$ n $\le$ 2¹⁰ , n even). The number n is the size of images, one image is represented internally by anxn pixel matrix (a^j_i) , where i is the row number andj is the column number. The pixel at the upper left corner is at row 0 and column 0.

The second line is a non-empty list of at most 32 words, separated by spaces. Valid words are the keywordsid, rot, sym, bhsym, bvsym,div and mix, or a keyword followed by ``-". Each keyword key designates an elementary transform (as defined by Figure 1), andkey- designates the inverse of transform key. For instance, rot- is the inverse of counter-clockwise 90^o rotation, that is clockwise 90^o rotation. Finally, the list k₁, k₂,..., k_p designates the compound transform $\phi$ =k₁ok₂o^...ok_p . For instance, ``bvsym rot-" is the transform that first performs clockwise 90^o rotation and then vertical symmetry on the lower half of the image.

$\epsfbox{p3510c.eps}$

Figure 1: Transformations of image (a^j_i) into image(b^j_i)

$\textstyle \parbox{.70\textwidth}{\begin{description}\vspace{.25in}\item[\te...... , a^{n/2+1}_{2k+1} , \cdots, a^{n-1}_{2k}, a^{n-1}_{2k+1}$.\end{description}}$ $\textstyle \parbox{.25\textwidth}{\begin{center}\mbox{}\epsfbox{p3510d.eps}\end{center}}$

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

Your program should output a single line whose contents is the minimal number m(m > 0) such that $\phi^{{m}}_{}$ is the identity. You may assume that, for all test input, you havem < 2³¹ .

Sample Input

2

256
rot- div rot div

256
bvsym div mix

Sample Output

8

63457题意：给你n*n的矩阵和几种操作，问重复几次后会得到原图思路：每个命令都是一个置换，操作序列就是置换的乘积，我们可以最终的处理出这个矩阵，一个结论：对于一个长度为l的循环A，当且仅当m是l的整数倍的时候A^m才是全等置换，所以答案是多个循环的最小公倍数#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1100;

int n, g[maxn][maxn], vis[maxn][maxn], save[maxn][maxn];
char str[maxn], tmp[maxn];

int gcd(int a, int b) {
	return b == 0 ? a : gcd(b, a % b);
}

int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}

void id() {
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			g[i][j] = save[i][j];
}

void rot(int flag) { 
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++) {
			if (flag)
				save[i][j] = g[n-j-1][i];	
			else save[n-j-1][i] = g[i][j];
		}
	id();	
}

void sym(int flag) {
	for (int i = 0; i < n; i++)	
		for (int j = 0; j < n; j++)
			save[i][j] = g[i][n-1-j];
	id();
}

void bhsym(int flag) {
	for (int i = 0; i < n/2; i++) 
		for (int j = 0; j < n; j++)
			save[i][j] = g[i][j];
	for (int i = n/2; i < n; i++) 
		for (int j = 0; j < n; j++)
			save[i][j] = g[i][n-1-j];
	id();
}

void bvsym(int flag) {
	for (int i = 0; i < n; i++) 
		for (int j = 0; j < n; j++) {
			if (i < n / 2) 
				save[i][j] = g[i][j];
			else save[i][j] = g[3 * n / 2 - 1 - i][j];
		}
	id();
}

void div(int flag) {
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++) {
			if (flag == 1) {
				if (i & 1) 
					save[i / 2 + n / 2][j] = g[i][j];
				else save[i / 2][j] = g[i][j];
			}
			else {
				if (i & 1)
					save[i][j] = g[i / 2 + n / 2][j];
				else save[i][j] = g[i / 2][j];
			}
		}
	id();
}

void mix(int flag) {
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++) {
			if ((i & 1) == 0) {
				if (flag) {
					if ((j & 1) == 0)
						save[i][j] = g[i][j / 2];
					else save[i][j] = g[i + 1][j / 2];
				}
				else {
					if ((j & 1) == 0)
						save[i][j / 2] = g[i][j];
					else save[i + 1][j / 2] = g[i][j];
				}
			}
			else {
				if (flag) {
					if ((j & 1) == 0)
						save[i][j] = g[i - 1][n / 2 + j / 2];
					else save[i][j] = g[i][n / 2 + j / 2];
				}
				else {
					if ((j & 1) == 0)
						save[i - 1][n / 2 + j / 2] = g[i][j];
					else save[i][n / 2 + j / 2] = g[i][j];

				}
			}
		}
	id();
}

void change(char *str) {
	int len = strlen(str);
	int flag = 1;
	if (str[0] == '-') {
		flag = 0;
		str++;
	}
	if (strcmp(str, "tor") == 0) rot(flag);
	else if (strcmp(str, "mys") == 0) sym(flag);
	else if (strcmp(str, "myshb") == 0) bhsym(flag);
	else if (strcmp(str, "mysvb") == 0) bvsym(flag);
	else if (strcmp(str, "vid") == 0) div(flag);
	else if (strcmp(str, "xim") == 0) mix(flag);
}

void tra() {
	int len = strlen(str);
	int sn = 0;
	for (int i = len - 1; i >= 0; i--) {
		if (str[i] == ' ') {
			tmp[sn] = '\0';
			change(tmp);
			sn = 0;
		}
		else {
			tmp[sn++] = str[i];
		}
	}
	tmp[sn] = '\0';
	change(tmp);
}

int solve() {
	int ans = 1;
	memset(vis, 0, sizeof(vis));
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			if (!vis[i][j]) {
				vis[i][j] = 1;
				int cnt = 1;
				int x = g[i][j] / n;
				int y = g[i][j] % n;
				while (!vis[x][y]) {
					cnt++;
					vis[x][y] = 1;
					int t = g[x][y] / n;
					y = g[x][y] % n;
					x = t;
				}
				ans = lcm(ans, cnt);
			}
		}
	}
	return ans;
}

void init() {
	scanf("%d", &n);
	getchar();
	gets(str);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			g[i][j] = i * n + j;
		}	
	}
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		init();
		tra();
		printf("%d\n", solve());
		if (t) 
			printf("\n");
	}
	return 0;
}

UVA - 1156 Pixel Shuffle （置换+模拟）

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