给定一个 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 到右下角 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;
}
- A.
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][1]; - B.
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; - C.
dp[i][j] = min(dp[i - 1][j], dp[i][j]) + grid[i][j]; - D.
dp[i][j] = min(dp[i][j], dp[i][j - 1]) + grid[i][j];