95. 费解的开关
题目链接:
https://www.acwing.com/problem/content/97/
题解:
前一行的状态可以决定后一行的按法,因为一个开关按两次等于没按,所以第一行的状态确定了,第二行就必须那么按,我们可以枚举第一行的按法,然后进行模拟,因为一行有5个框框,就有32种按法,用5位的二进制表示,例如00001表示,按第5个(1表示按,0表示不按)
AC代码:
public class NO95 {
static int t;
static int N = 6;
static char[][] arr = new char[N][N];
static char[][] temp = new char[N][N];
static int[] dx = {0, -1, 1, 0, 0};
static int[] dy = {0, 0, 0, -1, 1};
static void turn(int x, int y) {
for (int i = 0; i < 5; i++) {
int nx = x + dx[i];
int ny = y + dy[i];
if (nx < 0 || nx >= 5 || ny < 0 || ny >= 5) continue;
arr[nx][ny] ^= 1;
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
t = sc.nextInt();
while (t-- > 0) {
for (int i = 0; i < 5; i++) {
arr[i] = sc.next().toCharArray();
}
int res = 10;
for (int i = 0; i < 32; i++) { // 枚举第1行按法
int step = 0;
for (int j = 0; j < 5; j++) temp[j] = Arrays.copyOf(arr[j], arr[j].length);
for (int j = 0; j < 5; j++) { // 1代表按j,0代表不按j
if ((i >> j & 1) == 1) {
step++;
turn(0, j);
}
}
for (int j = 0; j < 4; j++) {
for (int k = 0; k < 5; k++) {
if (arr[j][k] == ‘0‘) {
step++;
turn(j + 1, k);
}
}
}
boolean flag = true;
for (int j = 0; j < 5; j++) {
if (arr[4][j] == ‘0‘) {
flag = false;
break;
}
}
if (flag) {
if (step < res) res = step;
}
for (int j = 0; j < 5; j++) arr[j] = Arrays.copyOf(temp[j], temp[j].length);
}
if (res <= 6) System.out.println(res);
else System.out.println(-1);
}
}
}