先特判掉一整行和一整列顶满的状态。
然后考虑极大化矩形面积，一定是边框压着关键点最优。
所以枚举每个点作为开头，以 $y$ 坐标排序，并向右扫更新上下边界。
如果 $x_j \lt x_i$ 就更新 $up$，否则更新 $down$。

说句闲话，这题貌似是 DP 题，但是写成了一个类似于悬线法的东西……

#include <bits/stdc++.h>
using namespace std;
const int N = 5015;
struct node {
    int x, y;
} a[N];

bool cmp1(node a, node b) {
    return a.x < b.x;
}

int k, n, m;
bool cmp2(node a, node b) {
    return a.y < b.y;
}

long long ans = 0;

int main() {
    scanf("%d%d%d", &m, &n, &k);
    for (int i = 1; i <= k; i++) scanf("%d%d", &a[i].y, &a[i].x);
    a[++k] = (node){0, 0}, a[++k] = (node){n, 0}, a[++k] = (node){0, m}, a[++k] = (node){n, m};
    sort(a + 1, a + 1 + k, cmp1);
    for (int i = 1; i < k; i++) ans = max(ans, m * 1ll * (a[i + 1].x - a[i].x));
    sort(a + 1, a + 1 + k, cmp2);
    for (int i = 1; i < k; i++) ans = max(ans, n * 1ll * (a[i + 1].y - a[i].y));

    for (int i = 1; i <= k; i++) {
        int up = 0, dn = n;
        for (int j = i + 1; j <= k; j++) {
            if (a[j].y == a[i].y) continue;
            ans = max(ans, (a[j].y - a[i].y) * 1ll * (dn - up));
            if (a[j].x < a[i].x) up = max(up, a[j].x);
            if (a[j].x >= a[i].x) dn = min(dn, a[j].x);
        }
    }

    for (int i = k; i >= 1; i--) {
        int up = 0, dn = n;
        for (int j = i - 1; j >= 1; j--) {
            if (a[j].y == a[i].y) continue;
            ans = max(ans, (a[i].y - a[j].y) * 1ll * (dn - up));    //what??? a[j].y - a[i].y ??
            if (a[j].x < a[i].x) up = max(up, a[j].x);
            if (a[j].x >= a[i].x) dn = min(dn, a[j].x);
        }
    }
    printf("%lld\n", ans);
    return 0;
}