这道题的c++代码是什么？

在 C++ 中解决这个问题，你可以使用二维动态规划（Dynamic Programming）和邻接矩阵来存储奶牛的位置和状态。以下是简单的代码示例：

#include <iostream>
#include <vector>

using namespace std;

int numNeighbours(vector<vector<int>>& grid, int x, int y) {
    int count = 0;
    for (int dx = -1; dx <= 1; dx++) {
        for (int dy = -1; dy <= 1; dy++) {
            if (dx == 0 && dy == 0) continue; // Skip the current cell
            int nx = x + dx, ny = y + dy;
            if (nx >= 0 && nx < grid.size() && ny >= 0 && ny < grid[0].size()) {
                count += grid[nx][ny];
            }
        }
    }
    return count;
}

int minExtraCows(vector<vector<int>>& grid) {
    int cows = 0;
    vector<vector<int>> dp(grid.size(), vector<int>(grid[0].size(), 0));
    for (int i = 0; i < grid.size(); i++) {
        for (int j = 0; j < grid[0].size(); j++) {
            if (grid[i][j] != 0) { // A cow is already there
                dp[i][j] = numNeighbours(grid, i, j);
            } else {
                dp[i][j] = 1; // A free space initially has one neighbour
            }
            if (dp[i][j] > 1) { // Need to add a cow to make it non-comfortable
                cows++;
            }
        }
    }
    return cows;
}

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

    vector<int> result(n);
    for (int i = 0; i < n; i++) {
        result[i] = minExtraCows(grid);
    }

    for (int extra : result) {
        cout << extra << "\n";
    }

    return 0;
}

这段代码首先定义了计算单个位置周围奶牛数量的函数 `numNeighbours`，接着计算整个地图上每个位置至少需要添加多少只奶牛的 `minExtraCows` 函数。主函数读取输入并输出结果。