576. Out of Boundary Paths

最新推荐文章于 2020-11-25 15:34:28 发布

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最新推荐文章于 2020-11-25 15:34:28 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/jxr041100/p/8115093.html

本文探讨了一个m*n网格中，从指定起点出发，在限定步数内，球能以多少种方式出界的算法问题。提供了两种解决方案：一种采用递归深度优先搜索配合记忆化的方法；另一种则使用动态规划技术来解决该问题。

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:

class Solution {
public:
    int findPaths(int m, int n, int N, int i, int j) {
        vector<vector<vector<int>>> dp(N + 1, vector<vector<int>>(m, vector<int>(n, -1)));       
        return dfs(m,n,N,i,j,dp);       
    }
private:   
    int dfs(int m,int n,int N,int i,int j,vector<vector<vector<int>>> &dp)
    {
        if(N==0) return 0;
        int res = 0;
        if(dp[N][i][j]!=-1) return dp[N][i][j];
        vector<pair<int,int>> dirs = {{-1,0},{1,0},{0,-1},{0,1}};
        for(auto dir:dirs)
        {
            int next_i = i+dir.first;
            int next_j = j+dir.second;
            if(next_i<0 || next_j <0 || next_i>=m || next_j >=n) //out of boundary
                res = res+1;
            else 
                res+=dfs(m,n,N-1,next_i,next_j,dp);
            res = res%1000000007;
        }
        res = res%1000000007; 
        dp[N][i][j] =  res;
        return res;
    }
};

动态规划的解法

class Solution {
public:
    int findPaths(int m, int n, int N, int i, int j) {
        vector<vector<vector<int>>> dp(N + 1, vector<vector<int>>(m, vector<int>(n, 0)));
        for (int k = 1; k <= N; ++k) {
            for (int x = 0; x < m; ++x) {
                for (int y = 0; y < n; ++y) {
                    long long v1 = (x == 0) ? 1 : dp[k - 1][x - 1][y];
                    long long v2 = (x == m - 1) ? 1 : dp[k - 1][x + 1][y];
                    long long v3 = (y == 0) ? 1 : dp[k - 1][x][y - 1];
                    long long v4 = (y == n - 1) ? 1 : dp[k - 1][x][y + 1];
                    dp[k][x][y] = (v1 + v2 + v3 + v4) % 1000000007;
                }
            }
        } 
        return dp[N][i][j];
    }
};