https://leetcode.com/problems/minimum-falling-path-sum/description/

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

用 d[i][j] 表示走到 (i, j) 处最小的和。
状态转移方程：
d[i][j] = min(d[i-1][j-1], d[i-1][j], d[i-1][j+1]) + A[i][j]
初始值：
d[0][j] = A[0][j]

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        const int INF = 1000000000;
        int n = A.size();
        int d[105][105];
        for (int j = 0; j < n; j++) {
            d[0][j] = A[0][j];
        }
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < n; j++) {
                d[i][j] = INF;
                if (j-1 >= 0) d[i][j] = min(d[i-1][j-1] + A[i][j], d[i][j]);
                              d[i][j] = min(d[i-1][j] + A[i][j], d[i][j]);
                if (j+1 < n)  d[i][j] = min(d[i-1][j+1] + A[i][j], d[i][j]);
            }
        }
        
        int ans = d[n-1][0];
        for (int j = 1; j < n; j++) {
            ans = min(ans, d[n-1][j]);
        }
        return ans;
    }
};