[leetcode]576. Out of Boundary Paths

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本文介绍了一个算法问题，即在一个m x n的网格中，从指定起点出发，在不超过N次移动的情况下，计算球可以穿越网格边界的路径总数。文章提供了一个C++实现方案，通过动态规划方法递推计算出所有可能路径的数量。

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题目链接：https://leetcode.com/problems/out-of-boundary-paths/#/description

There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 109 + 7.

Example 1:

Input:m = 2, n = 2, N = 2, i = 0, j = 0
Output: 6
Explanation:

Example 2:

Input:m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

Note:

Once you move the ball out of boundary, you cannot move it back.
The length and height of the grid is in range [1,50].
N is in range [0,50].

There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 109 + 7.

Example 1:
Input:m = 2, n = 2, N = 2, i = 0, j = 0
Output: 6
Explanation:

Example 2:
Input:m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

Note:
Once you move the ball out of boundary, you cannot move it back.
The length and height of the grid is in range [1,50].
N is in range [0,50].

The number of paths for N moves is the sum of paths for N - 1 moves from the adjacent cells. If an adjacent cell is out of the border, the number of paths is 1.

class Solution{
public:
    int findPaths(int m,int n,int N,int i,int j)
    {   //here can only use uint ,not int 
        uint dp[51][50][50]={};
        for(auto Ni=1;Ni<=N;Ni++)
        {
            for(auto mi=0;mi<m;mi++)
            {
                for(auto ni=0;ni<n;ni++)
                {
                    dp[Ni][mi][ni]=((mi==0?1:dp[Ni-1][mi-1][ni])+(mi==m-1?1:dp[Ni-1][mi+1][ni])+
                                    (ni==0?1:dp[Ni-1][mi][ni-1])+(ni==n-1?1:dp[Ni-1][mi][ni+1]))%1000000007;
                }
            }
        }
        return dp[N][i][j];
    }
};