LWC 52：688. Knight Probability in Chessboard
LWC 52：688. Knight Probability in Chessboard
LWC 52：688. Knight Probability in Chessboard
传送门：688. Knight Probability in Chessboard
Problem:
On an NxN chessboard, a knight starts at the r-th row and c-th column and attempts to make exactly K moves. The rows and columns are 0 indexed, so the top-left square is (0, 0), and the bottom-right square is (N-1, N-1). A chess knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.
Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there. The knight continues moving until it has made exactly K moves or has moved off the chessboard. Return the probability that the knight remains on the board after it has stopped moving.
Example:
Input: 3, 2, 0, 0 Output: 0.0625 Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board. From each of those positions, there are also two moves that will keep the knight on the board. The total probability the knight stays on the board is 0.0625.
Note:
N will be between 1 and 25.
K will be between 0 and 100.
The knight always initially starts on the board.
思路： 一开始采用BFS来遍历所有情况，但在OJ上内存溢出且超时了，主要原因在于BFS把所有的状态都跑一边，而此处K相对较大，N却较小，所以骑士在经过k步之后，很有可能在图中经过相同的点，只是对应的k不同罢了。
初始代码如下：
int[][] dir = {{2, 1},{2, -1},{-2, -1},{-2, 1},{1, 2},{1, -2},{-1, 2},{-1, -2}};
class Pair{
int id;
double pro;
Pair(int id, double pro){
this.id = id;
this.pro = pro;
}
}
public double knightProbability(int N, int K, int r, int c) {
Queue<Pair> queue = new ArrayDeque<>();
queue.offer(new Pair(r * N + c, 1));
int turn = 0;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; ++i) {
Pair p = queue.poll();
int cnt = 0;
List<Integer> tmp = new ArrayList<>();
for (int[] d : dir) {
int nx = d[0] + p.id / N;
int ny = d[1] + p.id % N;
if (check(nx, ny, N)) {
cnt ++;
tmp.add(nx * N + ny);
}
}
for (int j = 0; j < tmp.size(); ++j) {
queue.offer(new Pair(tmp.get(j), p.pro * cnt / 8));
}
}
turn ++;
if (turn == K) break;
}
double ans = 0;
int size = queue.size();
while (!queue.isEmpty()) {
ans += (queue.poll().pro / size);
}
return ans;
}
boolean check(int i, int j, int N) {
return i >= 0 && i < N && j >= 0 && j < N;
}这里遇到重复子问题的多次计算，定义问题为f(i, j, k)表示当前坐标(i, j)下，骑士k次移动后的概率。如何划分子问题呢？因为它有八个方向可以移动，所以有：f(i + d[x][0], j + d[x][1], k - 1), x = 1,2,...,8，表示移动到八个位置的其中一个后，原问题变成了如上子问题。
所以，我们有递归+记忆化的手段，代码如下：
int[][] dir = {{2, 1},{2, -1},{-2, -1},{-2, 1},{1, 2},{1, -2},{-1, 2},{-1, -2}};
double[][][] mem = new double[102][32][32];
public double f(int N, int K, int R, int C) {
if (K == 0) {
return 1;
}
if (mem[K][R][C] > 0) return mem[K][R][C];
double res = 0.0;
for (int[] d : dir) {
int nx = R + d[0];
int ny = C + d[1];
if (check(nx, ny, N)) {
res += 1 / 8.0 * f(N, K - 1, nx, ny);
}
}
return mem[K][R][C] = res;
}
public void fill(double[][][] mem) {
for (int i = 0; i < mem.length; ++i) {
for (int j = 0; j < mem[i].length; ++j) {
for (int k = 0; k < mem[i][j].length; ++k) {
mem[i][j][k] = -1;
}
}
}
}
boolean check(int i, int j, int N) {
return i >= 0 && i < N && j >= 0 && j < N;
}
public double knightProbability(int N, int K, int r, int c) {
fill(mem);
return f(N, K, r, c);
}当然，你可以直接把它改成自底向上的迭代形式（动态规划）。代码如下：
int[][] dir = {{2, 1},{2, -1},{-2, -1},{-2, 1},{1, 2},{1, -2},{-1, 2},{-1, -2}};
boolean check(int i, int j, int N) {
return i >= 0 && i < N && j >= 0 && j < N;
}
public double knightProbability(int N, int K, int r, int c) {
double[][][] mem = new double[102][32][32];
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
mem[0][i][j] = 1;
}
}
for (int k = 1; k <= K; ++k) {
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
for (int[] d : dir) {
int nx = i + d[0];
int ny = j + d[1];
if (check(nx, ny, N)) {
mem[k][i][j] += 1 / 8.0 * mem[k - 1][nx][ny];
}
}
}
}
}
return mem[K][r][c];
}