codeforces248(div1) B Nanami's Digital Board

q次询问，每次询问可以对矩阵某一个值改变(0变1，1变0) 或者是查询子矩阵的最大面积，要求这个这个点在所求子矩阵的边界上，且子矩阵各店中全为1

用up[i][j]表示(i,j)这个点向上能走到的最长高度  若(i,j)为0 则up[i][j]值为0

同理，维护down,left, right数组

则每次查询时，从up[i][j]枚举至1作为子矩阵的高度，然后途中分别向左右扩展。若up[i][j - 1] >= up[i][j]，则可向左扩展一个单位，答案为(r - l - 1) * 高度

同理，四个方向分别枚举

//#pragma comment(linker, "/STACK:102400000,102400000")
//HEAD
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>

#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <cstdlib>

using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//STL
#define PB push_back
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;


#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
#define SZ(V) (int)V.size()
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
typedef vector <int> VI;
typedef unsigned long long ULL;
const double eps = 1e-10;
const LL MOD = 1e9 + 7;
const int maxn = 1010;

int ipt[maxn][maxn];
int up[maxn][maxn], dwn[maxn][maxn], rht[maxn][maxn], lft[maxn][maxn];
int len[maxn], n, m;

void update_col(int y)
{
    FE(i, 1, n)
        if (ipt[i][y])
            up[i][y] = up[i - 1][y] + 1;
        else    up[i][y] = 0;
    FED(i, n, 1)
        if (ipt[i][y])
            dwn[i][y] = dwn[i + 1][y] + 1;
        else
            dwn[i][y] = 0;
}

void update_row(int x)
{
    FE(j, 1, m)
        if (ipt[x][j])
            lft[x][j] = lft[x][j - 1] + 1;
        else    lft[x][j] = 0;
    FED(j, m, 1)
        if (ipt[x][j])
            rht[x][j] = rht[x][j + 1] + 1;
        else    rht[x][j] = 0;
}

int solve(int sta, int hei, int con)
{
    int lm = sta, rm = sta;
    int ans = 0;
    for (int h = hei; h >= 1; h--)
    {
        while (lm >= 1 && len[lm] >= h)
            lm--;
        while (rm <= con && len[rm] >= h)
            rm++;
        ans = max(ans, h * (rm - lm - 1));
    }
    return ans;
}

int main()
{
    //freopen("0.txt", "r", stdin);
    int q, x, y, op;
    cin >> n >> m >> q;
    FE(i, 1, n)
        FE(j, 1, m)
            RI(ipt[i][j]);
    FE(i, 1, n)
        update_row(i);
    FE(j, 1, m)
        update_col(j);
    while (q--)
    {
        RIII(op, x, y);
        if (op == 1)
        {
            ipt[x][y] ^= 1;
            update_row(x);
            update_col(y);
//            cout << "UP  " << endl;
//            FE(i, 1, n) {
//                FE(j, 1, m)
//                    cout << up[i][j] << &#39; &#39;;
//                    cout <<endl;
//            }
//            cout << "----" << endl;
//            cout << "right  " << endl;
//            FE(i, 1, n) {
//                FE(j, 1, m)
//                    cout << rht[i][j] << &#39; &#39;;
//                    cout <<endl;
//            }
//            cout << "----" << endl;
        }
        else
        {
            int ans = 0;
            FE(j, 1, m) len[j] = up[x][j];
            ans = max(ans, solve(y, len[y], m));
            FE(j, 1, m) len[j] = dwn[x][j];
            ans = max(ans, solve(y, len[y], m));
            FE(i, 1, n) len[i] = lft[i][y];
            ans = max(ans, solve(x, len[x], n));
            FE(i, 1, n) len[i] = rht[i][y];
            ans = max(ans, solve(x, len[x], n));
            WI(ans);
        }
    }
    return 0;
}

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