Solution
#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar()) s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x) { if (!x) { putchar('0'); return; } char F[40]; 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 dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int N = 1e3 + 7;
int mp[N][N], n, m;
int c[N][N]; //树状数组
void add(int x, int val, int id) {
for (; x <= n; x += lowbit(x)) c[id][x] += val;
}
int query(int x, int id) {
int res = 0;
for (; x; x -= lowbit(x)) res += c[id][x];
return res;
}
int main() {
n = read(), m = read();
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j) {
mp[i][j] = read();
if (mp[i][j]) add(j, 1, i);
}
while (m--) {
int op = read();
if (op & 1) {
int x = read(), y = read();
if (mp[x][y]) { add(y, -1, x); mp[x][y] = 0; }
else { add(y, 1, x); mp[x][y] = 1; }
}
else {
int x1 = read(), y1 = read(), x2 = read(), y2 = read();
if (x1 > x2) swap(x1, x2);
if (y1 > y2) swap(y1, y2); //确保矩阵左上角和右小角点正确
int ans = 0;
for (int i = x1; i <= x2; ++i)
ans += query(y2, i) - query(y1 - 1, i);
print(ans), putchar(10);
}
}
return 0;
}