能否用c++语言来写上述问题

当然可以，以下是同样功能的 C++ 代码实现，供你参考：

#include <iostream>
#include <queue>
#include <unordered_set>
#include <string>

using namespace std;

typedef pair<int, int> pii;

// 哈希函数
struct HashFunc {
    size_t operator()(const pii& p) const {
        return hash<int>()(p.first) ^ hash<int>()(p.second);
    }
};

// 判断一个位置是否合法
bool is_valid(int x, int y, const unordered_set<pii, HashFunc>& obstacles) {
    if (x < -100 || x > 100 || y < -100 || y > 100) return false;
    if (obstacles.count({x, y})) return false;
    return true;
}

// 广搜
int bfs(pii start, pii end, const unordered_set<pii, HashFunc>& obstacles) {
    vector<pii> dirs = {{1, 2}, {2, 1}, {-1, 2}, {-2, 1}, {1, -2}, {2, -1}, {-1, -2}, {-2, -1}};
    queue<pair<pii, int>> q;
    unordered_set<pii, HashFunc> visited;
    q.push({start, 0});
    visited.insert(start);
    while (!q.empty()) {
        auto [pos, steps] = q.front();
        q.pop();
        if (pos == end) return steps;
        for (auto [dx, dy] : dirs) {
            int new_x = pos.first + dx, new_y = pos.second + dy;
            if (is_valid(new_x, new_y, obstacles)) {
                pii new_pos = {new_x, new_y};
                if (!visited.count(new_pos)) {
                    q.push({new_pos, steps + 1});
                    visited.insert(new_pos);
                }
            }
        }
    }
    return -1;
}

int main() {
    int x_final, y_final;
    cin >> x_final >> y_final;
    int n;
    cin >> n;
    unordered_set<pii, HashFunc> obstacles;
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        obstacles.insert({x, y});
    }
    pii start = {0, 0};
    pii end = {x_final, y_final};
    int res = bfs(start, end, obstacles);
    if (res == -1) {
        cout << "fail" << endl;
    } else {
        cout << res << endl;
    }
    return 0;
}

希望这份代码能够对你有所帮助！