注意子矩阵中，$x_1,y_1,x_2,y_2$ 的坐标是在线上，那么实际单行或单列方格数量会少1，又因为数据取值可以取到0，可以考虑 $x_1++, y_1++$。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1005, INF = 0x3f3f3f3f, MOD = 1E9 + 7;
int n = 1000, k, st[N][N], ans;
void solve1() {
    cin >> k;
    int x1, y1, x2, y2;
    while (k--) {
        cin >> x1 >> y1 >> x2 >> y2;
        x1++, y1++;
        for (int i = x1; i <= x2; i++)
            for (int j = y1; j <= y2; j++)
                st[i][j] = 1;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            ans += st[i][j];
            // cout << st[i][j] << " \n"[j == n];
        }
    cout << ans << endl;
}
int solve2() {
    cin >> k;
    int x1, y1, x2, y2;
    while (k--) {
        cin >> x1 >> y1 >> x2 >> y2;
        x1++, y1++;
        st[x1][y1]++;
        st[x1][y2 + 1]--;
        st[x2 + 1][y1]--;
        st[x2 + 1][y2 + 1]++;
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            st[i][j] += st[i - 1][j] + st[i][j - 1] - st[i - 1][j - 1];
            if (st[i][j] > 0)
                ans++;
        }
    cout << ans << endl;
    return 0;
}
int main() {
    solve2();
}