Bzoj 3039: Jade Toad Hall (suspension line method)

Maximum sub-matrix ... Suspension Line Method: Time complexity O (nm) http://blog.csdn.net/wzq_qwq/article/details/47167707

The method of suspension is to record the maximum length of an h extension (hanging line), L, R to the left to the right to extend the maximum length, and then by recursion to get.

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#include <bits/stdc++.h>using namespace std;#define OK (c) ((c) = = ' F ' | | (c) = = ' R ')const int MAXN = 1009; int H[MAXN][MAXN], L[MAXN][MAXN], R[MAXN][MAXN], N, M;bool F[MAXN][MAXN];void Read () {cin >> N >> M;for (int i = 0; i < N; i++)For (int j = 0; J < M; J + +) {char C = getchar ();For (;!ok (c); c = GetChar ());F[i][j] = c = = ' F ';    }}int main () {Read ();for (int i = 0; i < N; i++) {l[i][0] = f[i][0];For (int j = 1; j < M; J + +)L[i][j] = f[i][j]? L[i][j-1] + 1:0;r[i][m-1] = f[i][m-1];For (int j = M-2; ~j; j--)R[i][j] = f[i][j]? R[i][j + 1] + 1:0;}memset (h, 0, sizeof h);for (int i = 1; i < N; i++)For (int j = 0; J < M; J + +) if (F[i][j] && f[i-1][j]) {H[i][j] = h[i-1][j] + 1;L[i][j] = min (L[i][j], l[i-1][j]);R[i][j] = min (R[i][j], r[i-1][j]);}int ans = 0;for (int i = 0; i < N; i++)For (int j = 0; J < M; J + +) if (F[i][j])ans = max (ans, (h[i][j] + 1) * (L[i][j] + r[i][j]-1));cout << 3 * ans << "\ n";return 0;}

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Bzoj 3039: Jade Toad Hall (suspension line method)