给定一个 $m \times n$ 的二维网格 grid，每个格子中有一个非负整数。请找出一条从左上角 $(0, 0)$ 到右下角 $(m-1, n-1)$ 的路径，使得路径上的数字总和最小。每次只能向右或向下移动。横线处应该填入的是（ ）

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int minPathSum(vector<vector<int>>& grid) {
    int m = grid.size();
    int n = grid[0].size();

    vector<vector<int>> dp(m, vector<int>(n, 0));
    dp[0][0] = grid[0][0];
    for (int j = 1; j < n; j++) {
        dp[0][j] = dp[0][j - 1] + grid[0][j];
    }
    for (int i = 1; i < m; i++) {
        dp[i][0] = dp[i - 1][0] + grid[i][0];
    }
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            ————————————————————————————————
        }
    }
    return dp[m - 1][n - 1];
}

int main() {
    int m, n;
    cin >> m >> n;
    vector<vector<int>> grid(m, vector<int>(n));
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            cin >> grid[i][j];
        }
    }
    int result = minPathSum(grid);
    cout << result << endl;

    return 0;
}

{{ select(1) }}

• dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][1];
• dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
• dp[i][j] = min(dp[i - 1][j], dp[i][j]) + grid[i][j];
• dp[i][j] = min(dp[i][j], dp[i][j - 1]) + grid[i][j];