[LeetCode] 688. Knight Probability in Chessboard
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[LeetCode] 688. Knight Probability in Chessboard
题目描述
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.
分析
每次计算每步当前所在格子的安全的可以跳跃的格子进行统计,最后得出在初始情况下安全的跳的方法总数,除以K步的总共跳的情况,即可得出概率。
记录下对于每步每个格子可以安全跳跃的格子的数量,这样可以避免在子问题中进行重复的计算。
class Solution {public: double knightProbability(int N, int K, int r, int c) { vector<vector<vector<double>>> dp(K + 1, vector<vector<double>>(N, vector<double>(N, -1.0))); return totalJump(dp, N, K, r, c) / pow(8, K); } double totalJump(vector<vector<vector<double>>> &dp, int n, int k, int r, int c) { if (r < 0 || r >= n || c < 0 || c >= n) return 0.0; if (k == 0) return 1.0; if (dp[k][r][c] != -1.0) return dp[k][r][c]; dp[k][r][c] = 0.0; for (int i = -2; i <= 2; i++) { for (int j = -2; j <= 2; j++) { if (abs(i) + abs(j) == 3) { dp[k][r][c] += totalJump(dp, n, k - 1, r + i, c + j); } } } return dp[k][r][c]; }};
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