108th LeetCode Weekly Contest Minimum Falling Path Sum

Given a square array of integers A, we want the minimum sum of a falling path through A.

A falling path starts at any element in the first row, and chooses one element from each row.  The next row's choice must be in a column that is different from the previous row's column by at most one.

Example 1:

Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: 12
Explanation: 
The possible falling paths are:

• [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
• [2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
• [3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]

The falling path with the smallest sum is [1,4,7], so the answer is 12.

Note:

1. 1 <= A.length == A[0].length <= 100
2. -100 <= A[i][j] <= 100

大概都知道求矩阵的最小和，这里不过是规则变了（需要是向下且不同行）

那个dp[i][j]可能从正上方，或者是右上，左上 + A[i][j]得来，考虑边界

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        if(A.size() == 0) return 0;
        int inf = 100000;
        int n = A.size();
        vector<vector<int>> dp(n + 10, vector<int>(n + 10, inf));
        for (int i = 0; i <= n; ++i) dp[0][i] = A[0][i];
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (j > 0) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + A[i][j]);
                if (j + 1 < n) dp[i][j] = min(dp[i][j], dp[i - 1][j + 1] + A[i][j]);
                dp[i][j] = min(dp[i][j], dp[i - 1][j] + A[i][j]);
            }
        }
        int ret = inf;
        for (int x : dp[n - 1]) ret = min(ret, x);
        return ret;
    }
};