You are an expert in writing code in many different languages. Your goal is to perform code completion for the following code, keeping in mind the rest of the code and the file meta data.
    Metadata details:
    filename: d.go
    path of file: ./repos/codeforces-go/leetcode/biweekly/66/d
    
    the code of the file until where you have to start completion: package main

/* 动态规划

先求正金字塔。

定义 $\textit{dp}[i][j]$ 表示金字塔顶端位于 $(i,j)$ 时的最大层数（$1$ 层也算）。如果顶端在 $(i,j)$ 的金字塔最大能有 $x$ 层，那么顶端在 $(i,j)$ 的金字塔也可以有 $x-1,x-2,\cdot,\1$ 层。由于要求区域内格子数目大于 $1$，最后统计答案的时候把 $1$ 层去掉，因此 $(i,j)$ 处可以有 $x-1$ 个金字塔。

我们从 $\textit{grid}$ 的最后一行开始往上递推。转移的策略是在金字塔上套一层**倒 V 型的「外壳」**。具体来说，设从 $(i,j)$ 出发向左下方向前进，求出能达到的连续 $1$ 的个数，根据 $\textit{dp}$ 的定义，这就是 $\textit{dp}[i+1][j-1]+1$；同理，向右下方向前进，能达到的连续 $1$ 的个数为 $\textit{dp}[i+1][j+1]+1$。

那么有

$$
\textit{dp}[i][
    
func countPyramids(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	dp := make([][]int, m)
	for i := range dp {
		dp[i] = make([]int, n)
	}

	f := func() {
		dp[m-1] = grid[m-1]
		for i := m - 2; i >= 0; i-- {
			dp[i][0] = grid[i][0]
			dp[i][n-1] = grid[i][n-1]
			for j := 1; j < n-1; j++ {
				if grid[i][j] == 0 {
					dp[i][j] = 0
				} else {
					dp[i][j] = min(min(dp[i+1][j-1], dp[i+1][j]), dp[i+1][j+1]) + 1
					ans += dp[i][j] - 1
				}
			}
		}
	}
	f()
	for i := 0; i < m/2; i++ {
		grid[i], grid[m-1-i] = grid[m-1-i], grid[i]
	}
	f()
	return
}