给定一个m×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（）
1 #include <iostream>
2 #include <vector>
3 #include <algorithm>
4 using namespace std;
5
6 int minPathSum(vector<vector<int>>& grid) {
7 int m = grid.size();
8 int n = grid[0].size();
9
10 vector<vector<int>> dp(m, vector<int>(n, 0));
11
12 dp[0][0] = grid[0][0];
13 for (int j = 1; j < n; j++) {
14 dp[0][j] = dp[0][j - 1] + grid[0][j];
15 }
16 for (int i = 1; i < m; i++) {
17 dp[i][0] = dp[i - 1][0] + grid[i][0];
18 }
19 for (int i = 1; i < m; i++) {
20 for (int j = 1; j < n; j++) {
21 ————————————————————————————————
22 }
23 }
24 return dp[m - 1][n - 1];
25 }
26
27 int main() {
28 int m, n;
29 cin >> m >> n;
30 vector<vector<int>> grid(m, vector<int>(n));
31 for (int i = 0; i < m; i++) {
32 for (int j = 0; j < n; j++) {
33 cin >> grid[i][j];
34 }
35 }
36 int result = minPathSum(grid);
37 cout << result << endl;
38
39 return 0;
40 }
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];