LA 7735 - Pocket Cube

LA 7735 - Pocket Cube

原题链接

题目类型：模拟

题意

给定一个二阶魔方每个面的状态，求能否在一步内将魔方还原。

分析

根据题目描述得出每个颜色在输入得数组中得下标。然后就是模拟魔方转动的过程并判断是否还原。思考后可以发现，只有六种转动的情况和一种不转的情况（顺时针转动前面的面等价于逆时针转动后面的面），所以以枚举每种情况进行判断。最好用纸折一个立方体，然后将每个颜色对应的下标写上去调参更方便、准确。

PS：题目数据貌似有问题，输入的 N 应该比实际的测试数据多一个，用 Java 不做处理是过不了的

代码

static int[] top = new int[4];
static int[] front = new int[4];
static int[] bottom = new int[4];
static int[] back = new int[4];
static int[] left = new int[4];
static int[] right = new int[4];
static boolean ok;

public static void solve() throws IOException {
    for (int i = 0; i < 4; i++) if (hasNext()) top[i] = nextInt();
    for (int i = 0; i < 4; i++) if (hasNext()) front[i] = nextInt();
    for (int i = 0; i < 4; i++) if (hasNext()) bottom[i] = nextInt();
    for (int i = 0; i < 4; i++) if (hasNext()) back[i] = nextInt();
    for (int i = 0; i < 4; i++) if (hasNext()) left[i] = nextInt();
    for (int i = 0; i < 4; i++) if (hasNext()) right[i] = nextInt();
    
    ok = isOk();
    
    // x-z
    for (int i = 0; i < 4; i++) 
    fun(top, 0, 2, front, 0, 2, bottom, 0, 2, back, 0, 2, left, 1, i);
    
    // y-z
    for (int i = 0; i < 4; i++) 
    fun(top, 0, 1, right, 0, 1, bottom, 3, 2, left, 0, 1, back, 3, i);
    
    // x-y
    for (int i = 0; i < 4; i++) 
    fun(front, 0, 1, right, 2, 0, back, 3, 2, left, 1, 3, top, 5, i);
    
    pw.println(ok ? "YES" : "NO");
}   

public static void fun(int[] a, int a1, int a2,  int[] b, int b1, int b2, int[] c, int c1, int c2, int[] d, int d1, int d2, int[] e, int k, int i) {

    int t1 = a[a1], t2 = a[a2];
    a[a1] = b[b1]; a[a2] = b[b2];
    b[b1] = c[c1]; b[b2] = c[c2];
    c[c1] = d[d1]; c[c2] = d[d2];
    d[d1] = t1;    d[d2] = t2;
    
    switch (k) {
        case 1: turn(e, 1, 3, 2, 0); break;
        case 3: turn(e, 2, 3, 1, 0); break;
        case 5: turn(e, 1, 0, 2, 3); break;
    }
    if (i == 0 || i == 2) ok |= isOk();
}

public static void turn(int[] e, int a1, int a2, int a3, int a4) {
    int t = e[a1];
    e[a1] = e[a2];
    e[a2] = e[a3];
    e[a3] = e[a4];
    e[a4] = t;
}

public static boolean isOk() {
    for (int i = 1; i < 4; i++) if (top[i] != top[i - 1]) return false;
    for (int i = 1; i < 4; i++) if (front[i] != front[i - 1]) return false;
    for (int i = 1; i < 4; i++) if (bottom[i] != bottom[i - 1]) return false;
    for (int i = 1; i < 4; i++) if (back[i] != back[i - 1]) return false;
    for (int i = 1; i < 4; i++) if (left[i] != left[i - 1]) return false;
    for (int i = 1; i < 4; i++) if (right[i] != right[i - 1]) return false;
    
    return true;
}