#include <bits/stdc++.h>
using namespace std;

int main(void)
{
    size_t d, n;
    vector<vector<size_t>> v(179, vector<size_t>(179));
    map<size_t, size_t> cnt;
    cin >> d >> n;
    for (size_t i = 0, x, y, t; i < n; ++i)
    {
        cin >> x >> y >> t;
        v[x + 20][y + 20] = t;
    }
    auto count = [](const vector<vector<size_t>> &v, const size_t &x, const size_t &y, const size_t &d) {
        size_t count = 0;
        for (size_t i = x - d; i <= x + d; ++i)
            for (size_t j = y - d; j <= y + d; ++j)
                try
                {
                    count += v.at(i).at(j);
                }
                catch (const std::exception)
                {
                }
        return count;
    };
    for (size_t i = 20; i < 149; i++)
        for (size_t j = 20; j < 149; j++)
            cnt[count(v, i, j, d)]++;
    cout << cnt.rbegin()->second << ' ' << cnt.rbegin()->first << '\n';
    // for_each(cnt.begin(), cnt.end(), [](auto &x) { cout << x.second << ' ' << x.first << '\n'; });
    return 0;
}

这段代码是可以通过的 但是，注意，当不填充时，即只开 129 x 129的数组，无论是判定范围还是异常处理，都会出现部分求和为0的情况，也就是答案为32 10 而不是 38 10 即

#include <bits/stdc++.h>
using namespace std;

int main(void)
{
    size_t d, n;
    vector<vector<size_t>> v(129, vector<size_t>(129)); // 范围
    map<size_t, size_t> cnt;
    cin >> d >> n;
    for (size_t i = 0, x, y, t; i < n; ++i)
    {
        cin >> x >> y >> t;
        v[x][y] = t; // 不做偏移
    }
    auto count = [](const vector<vector<size_t>> &v, const size_t &x, const size_t &y, const size_t &d) {
        size_t count = 0;
        for (size_t i = x - d; i <= x + d; ++i)
            for (size_t j = y - d; j <= y + d; ++j)
                try
                {
                    count += v.at(i).at(j);
                }
                catch (const std::exception)
                {
                }
        return count;
    };
    for (size_t i = 0; i < v.size(); i++)
        for (size_t j = 0; j < v[i].size(); j++) // 不做偏移
            cnt[count(v, i, j, d)]++;
    cout << cnt.rbegin()->second << ' ' << cnt.rbegin()->first << '\n';
    // for_each(cnt.begin(), cnt.end(), [](auto &x) { cout << x.second << ' ' << x.first << '\n'; });
    return 0;
}

测试用例 1 11 0 0 10 64 64 10 0 124 5 4 128 5 3 128 1 0 128 4 128 128 10 127 0 10 128 64 10 64 128 9 30 20 10 正确输出 38 10