解题思路

二维前缀和，模板题。

注意起始下标是0，边缘的不算，就能A了,就把价值算在下一个位置就行了

#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); while (ch < 48 || ch > 57) { if (ch == '-') w = -1; ch = getchar(); }    while (ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();    return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[200]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 5e3 + 7;
int a[N][N];

int main() {
    int n = read(), r = read();
    int x, y, v;
    for (int i = 1; i <= n; ++i)
        x = read(), y = read(), v = read(), a[x + 1][y + 1] = v;
    for (int i = 1; i < N; ++i)
        for (int j = 1; j < N; ++j)
            a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
    int ans = 0;
    for (int i = r; i < N; ++i)
        for (int j = r; j < N; ++j)
            ans = max(ans, a[i][j] - a[i - r][j] - a[i][j - r] + a[i - r][j - r]);
    write(ans), putchar(10);
    return 0;
}