[leetcode]Knight Probability in Chessboard

链接：https://leetcode.com/problems/knight-probability-in-chessboard/description/

Question

1. On an NxN chessboard, a knight starts at the r-th row and c-th column and attempts to make exactly Kmoves. 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.

Solution

class Solution {
public:
  int direction[8][2] = {
    {1,2},
    {-1,2},
    {1,-2},
    {-1,-2},
    {2,1},
    {-2,1},
    {2,-1},
    {-2,-1}
  };
  #define pro (double)1/8
  bool inMap(int x, int y, int N) { return x >= 0 && x < N && y >= 0 && y < N; }
  double knightProbability(int N, int K, int r, int c) {
    double dp[30][30][110] = {};
    memset(dp, 0, sizeof(dp));
    dp[r][c][0] = 1;
    for (int p = 1; p <= K; p++) {
      for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
          double sum = 0;
          for (int d = 0; d < 8; d++) {
            int x = i+direction[d][0];
            int y = j+direction[d][1];
            if (inMap(x, y, N)) {
              sum += dp[x][y][p-1]*pro;
            }
          }
          dp[i][j][p] = sum;
        }
      }
    }
    double ans = 0;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < N; j++)
        ans += dp[i][j][K];
    return ans;
  }
};

思路：这是一道动态规划题，dp[i][j][p]是指第p次到达(i,j)的概率，只需要统计所有可能的来源就可以了。