时间:2021-07-01 10:21:17 帮助过:4人阅读
const int maxn = 1100;int n, m, q;int ipt[maxn][maxn];int up[maxn][maxn], dwn[maxn][maxn], lft[maxn][maxn], rht[maxn][maxn];int len[maxn];void updaterow(int r){ FE(j, 1, m) lft[r][j] = (ipt[r][j] == 1 ? lft[r][j - 1] + 1 : 0); FED(j, m, 1) rht[r][j] = (ipt[r][j] == 1 ? rht[r][j + 1] + 1 : 0);}void updatecol(int c){ FE(i, 1, n) up[i][c] = (ipt[i][c] == 1 ? up[i - 1][c] + 1 : 0); FED(i, n, 1) dwn[i][c] = (ipt[i][c] == 1 ? dwn[i + 1][c] + 1 : 0);}int maxarea(int s, int len[], int thes){ int l = s, r = s, ret = 0; FED(i, len[s], 1) { while (l >= 1 && len[l] >= i) l--; while (r <= thes && len[r] >= i) r++; ret = max(ret, i * (r - l - 1)); } return ret;}int main(){ while (~RIII(n, m, q)) { FE(i, 1, n) FE(j, 1, m) RI(ipt[i][j]); FE(i, 1, n) updaterow(i); FE(j, 1, m) updatecol(j); REP(kase, q) { int op, x, y; RIII(op, x, y); if (op == 1) { ipt[x][y] ^= 1; updatecol(y); updaterow(x); } else { int ans = max(maxarea(y, up[x], m), maxarea(y, dwn[x], m)); FE(i, 1, n) len[i] = lft[i][y]; ans = max(ans, maxarea(x, len, n)); FE(i, 1, n) len[i] = rht[i][y]; ans = max(ans, maxarea(x, len, n)); WI(ans); } } } return 0;}