uva 1156 - Pixel Shuffle(模拟+置换)
题目链接：uva 1156 - Pixel Shuffle
题目大意：给定一个N*N的黑白位图，有7种操作，并且对应在指令后加上‘-’即为操作的逆，给定N和一系列操作，（从最后一个开始执行），问说这一套指令需要执行多少次才能形成循环。
解题思路：模拟指令执行后获得一个置换，分解成若干的循环，各个循环长度的最小公倍数即使答案。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = (1<<10) + 5;
int N, s[maxn][maxn], f[maxn][maxn];
int c, v[maxn];
const char sign[maxn][10] = {"id", "id-", "rot", "rot-", "sym", "sym-", "bhsym", "bhsym-", "bvsym",
"bvsym-", "div", "div-", "mix", "mix-" };
inline int get_sign (char* word) {
for (int i = 0; i < 14; i++)
if (strcmp(word, sign[i]) == 0)
return i;
return -1;
}
// 逆时针90
void rot (bool flag) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (flag)
f[i][j] = s[N-1-j][i];
else
f[N-1-j][i] = s[i][j];
}
}
memcpy(s, f, sizeof(f));
}
// 水平翻转
void sym () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
f[i][j] = s[i][N-1-j];
}
memcpy(s, f, sizeof(f));
}
// 下一半水平翻转
void bhsym () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i >= N/2)
f[i][j] = s[i][N-1-j];
else
f[i][j] = s[i][j];
}
}
memcpy(s, f, sizeof(f));
}
// 下一半垂直翻转
void bvsym () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i >= N/2)
f[i][j] = s[3 * N / 2 - i - 1][j];
else
f[i][j] = s[i][j];
}
}
memcpy(s, f, sizeof(f));
}
// 分成两份
void div (bool flag) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (flag) {
if (i&1)
f[i/2 + N/2][j] = s[i][j];
else
f[i/2][j] = s[i][j];
} else {
if (i&1)
f[i][j] = s[i/2 + N/2][j];
else
f[i][j] = s[i/2][j];
}
}
}
memcpy(s, f, sizeof(f));
}
void mix (bool flag) {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i&1) {
if (flag) {
if (j&1)
f[i][j] = s[i][N/2 + j/2];
else
f[i][j] = s[i-1][N/2 + j/2];
} else {
if (j&1)
f[i][N/2+j/2] = s[i][j];
else
f[i-1][N/2+j/2] = s[i][j];
}
} else {
if (flag) {
if (j&1)
f[i][j] = s[i+1][j/2];
else
f[i][j] = s[i][j/2];
} else {
if (j&1)
f[i+1][j/2] = s[i][j];
else
f[i][j/2] = s[i][j];
}
}
}
}
/* if (N&1) { for (int i = 0; i < N; i++) { f[i][n] = s[i][n]; f[n][i] = s[n][i]; } } */
memcpy(s, f, sizeof(f));
}
void put () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf("%d ", s[i][j]);
printf("\n");
}
}
void handle () {
for (int i = c-1; i >= 0; i--) {
switch (v[i]) {
case 0:
case 1:
break;
case 2:
rot(true);
break;
case 3:
rot(false);
break;
case 4:
case 5:
sym();
break;;
case 6:
case 7:
bhsym();
break;
case 8:
case 9:
bvsym();
break;
case 10:
div(true);
break;
case 11:
div(false);
break;
case 12:
mix(true);
break;
case 13:
mix(false);
break;
}
}
}
ll gcd (ll a, ll b) {
return b == 0 ? a : gcd(b, a % b);
}
ll lcm (ll a, ll b) {
return a / gcd (a, b) * b;
}
ll solve () {
char str[maxn], word[maxn];
gets(str);
int len = strlen(str), mv = 0;
c = 0;
for (int i = 0; i <= len; i++) {
if ((str[i] >= 'a' && str[i] <= 'z') || str[i] == '-')
word[mv++] = str[i];
else {
word[mv] = '\0';
v[c++] = get_sign(word);
mv = 0;
}
}
handle();
ll ret = 1;
memset(f, 0, sizeof(f));
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (f[i][j])
continue;
int x = s[i][j] / N;
int y = s[i][j] % N;
ll cnt = 0;
while (f[x][y] == 0) {
f[x][y] = 1;
cnt++;
int tmp = s[x][y];
x = tmp / N;
y = tmp % N;
}
ret = lcm(ret, cnt);
}
}
return ret;
}
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
scanf("%d%*c", &N);
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
s[i][j] = i * N + j;
printf("%lld\n", solve());
if (cas)
printf("\n");
}
return 0;
}