zoukankan      html  css  js  c++  java
  • !688. Knight Probability in Chessboard

    #week10

    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.

    题解:

     1 class Solution {
     2 public:
     3     double knightProbability(int N, int K, int r, int c) {
     4         vector<vector<vector<double>>> dp(K+1, vector<vector<double>>(N, vector<double>(N, -1.0)));
     5         return helper(dp, N, K, r, c)/pow(8, K);
     6     }
     7 private:
     8     double helper(vector<vector<vector<double>>>& dp, int N, int k, int r, int c) {
     9         // if out of board, return 0.0
    10         if (r < 0 || r >= N || c < 0 || c >= N) return 0.0;
    11         // when k = 0, no more move, so it's 100% safe
    12         if (k == 0) return 1.0;
    13         if (dp[k][r][c] != -1.0) return dp[k][r][c];
    14         dp[k][r][c] = 0.0;
    15         for (int i = -2; i <= 2; i++) {
    16             if (i == 0) continue;
    17             dp[k][r][c] += helper(dp, N, k-1, r+i, c+3-abs(i)) + helper(dp, N, k-1, r+i, c-(3-abs(i)));
    18         }      
    19         return dp[k][r][c];
    20     }
    21 };
  • 相关阅读:
    攻城狮在路上(肆)How tomcat works(一) 简单的web服务器
    攻城狮在路上(肆)How tomcat works(零) 前言说明
    font-face使用备忘
    subversion安装使用
    判断一个类到底是从哪个jar包中调用的工具类
    JavaScript实例
    Nginx(一)
    PHP面向对象(七)
    PHP面向对象(六)
    PHP面向对象(五)
  • 原文地址:https://www.cnblogs.com/iamxiaoyubei/p/8278263.html
Copyright © 2011-2022 走看看