LeetCode688. Knight Probability in Chessboard—

这篇具有很好参考价值的文章主要介绍了LeetCode688. Knight Probability in Chessboard——动态规划。希望对大家有所帮助。如果存在错误或未考虑完全的地方，请大家不吝赐教，您也可以点击"举报违法"按钮提交疑问。

一、题目

On an n x n chessboard, a knight starts at the cell (row, column) and attempts to make exactly k moves. The rows and columns are 0-indexed, so the top-left cell is (0, 0), and the bottom-right cell is (n - 1, n - 1).

A chess knight has eight possible moves it can make, as illustrated below. Each move is two cells in a cardinal direction, then one cell 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 1:

Input: n = 3, k = 2, row = 0, column = 0
Output: 0.06250
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.
Example 2:

Input: n = 1, k = 0, row = 0, column = 0
Output: 1.00000

Constraints:

1 <= n <= 25
0 <= k <= 100
0 <= row, column <= n - 1文章来源地址https://www.toymoban.com/news/detail-831519.html

二、题解

class Solution {
public:
    double knightProbability(int n, int k, int row, int column) {
        vector<vector<vector<double>>> dp(n,vector<vector<double>>(n,vector<double>(k+1,-1)));
        return f(n,row,column,k,dp);
    }
    //从(i,j)出发，还有k步要走，最后落在棋盘上的概率
    double f(int n,int i,int j,int k,vector<vector<vector<double>>>& dp){
        if(i < 0 || i >= n || j < 0 || j >= n) return 0;
        if(dp[i][j][k] != -1) return dp[i][j][k];
        double res = 0;
        if(k == 0) res = 1;
        else{
            res += (f(n,i-2,j+1,k-1,dp) / 8);
            res += (f(n,i-1,j+2,k-1,dp) / 8);
            res += (f(n,i+1,j+2,k-1,dp) / 8);
            res += (f(n,i+2,j+1,k-1,dp) / 8);
            res += (f(n,i+2,j-1,k-1,dp) / 8);
            res += (f(n,i+1,j-2,k-1,dp) / 8);
            res += (f(n,i-1,j-2,k-1,dp) / 8);
            res += (f(n,i-2,j-1,k-1,dp) / 8);
        }
        dp[i][j][k] = res;
        return res;
    }
};