题目链接:传送门

题目大意:

  求由F构成的最大子矩阵的面积。输出面积的三倍。

  1 ≤ N,M ≤ 1000。

思路:

  悬线法模板题。

#include <bits/stdc++.h>

using namespace std;
const int MAX_N = 1e3 + ; int N, M;
char mat[MAX_N][MAX_N];
int lef[MAX_N][MAX_N], rig[MAX_N][MAX_N], up[MAX_N][MAX_N]; void init()
{
for (int i = ; i <= N; i++) {
for (int j = ; j <= M; j++)
if (mat[i][j] == 'F')
lef[i][j] = lef[i][j-] + ;
else
lef[i][j] = ;
for (int j = M; j >= ; j--)
if (mat[i][j] == 'F')
rig[i][j] = rig[i][j+] + ;
else
rig[i][j] = ;
} } void dp()
{
int ans = ;
for (int i = ; i <= N; i++) {
for (int j = ; j <= M; j++) {
if (mat[i][j] == 'F') {
if (i > && mat[i-][j] == 'F') {
up[i][j] = up[i-][j]+;
lef[i][j] = min(lef[i][j], lef[i-][j]);
rig[i][j] = min(rig[i][j], rig[i-][j]);
}
else
up[i][j] = ;
}
else
up[i][j] = ;
int len = rig[i][j] + lef[i][j] - ;
int high = up[i][j];
int a = min(len, high);
ans = max(ans, a*a);
ans = max(ans, len*high);
}
}
cout << ans* << endl;
} int main()
{
cin >> N >> M;
for (int i = ; i <= N; i++)
for (int j = ; j <= M; j++)
cin >> mat[i][j];
init();
dp(); return ;
}
05-08 14:52