hdu 1505 City Game dp

题意：最大F子矩阵 * 3

每行处理出里（i，j）最左最有的不小于mat【i】【j】的数组，然后更新答案

#include <bits/stdc++.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
typedef long long ll;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 1010;

int mat[N][N];
char t[N][N];
int l[N], r[N];
int n, m;

int main()
{
    int tot;
    for( scanf("%d", &tot); tot--; ) {
        scanf("%d%d", &n, &m);
        //getchar();
        memset( mat, 0, sizeof( mat ) );
        for( int i = 1; i <= n; ++i ) {
            for( int j = 1; j <= m; ++j )
                //scanf("%*c%c", &t[i][j]);
                cin >> t[i][j];
        }
        for( int j = 1; j <= m; ++j ) {
            for( int i = 1; i <= n; ++i ) {
                if( t[i][j] == 'F' )
                    mat[i][j] = mat[i-1][j] + 1;
            }
        }
        int ans = -inf;
        for( int i = 1; i <= n; ++i ) {
            l[1] = 1, r[m] = m;
            for( int j = 2; j <= m; ++j ) {
                if( !mat[i][j] )
                    continue;
                int p = j;
                while( p > 1 && mat[i][p-1] >= mat[i][j] )
                    p = l[p-1];
                l[j] = p;
            }
            for( int j = m-1; j >= 1; --j ) {
                if( !mat[i][j] )
                    continue;
                int p = j;
                while( p < m && mat[i][p+1] >= mat[i][j] )
                    p = r[p+1];
                r[j] = p;
            }
            for( int j = 1; j <= m; ++j ) {
                ans = max( ans, mat[i][j] * (r[j] - l[j] + 1 ) );
            }
        }
        printf("%d\n", ans * 3);
    }
    return 0;
}