hdu1505

hdu1506的加强版， 如果要做这题，还是先去做1505吧

这一题，其实是对每一行做1505的那种dp，然后取最大值就行了。

#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <time.h>
#include <math.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <bitset>
#include <algorithm>
#include <iostream>
#include <string>
#include <functional>
#include <unordered_map>
typedef __int64 LL;
const int INF = 999999999;


const int N = 1000 + 10;
int a[N][N];
int sum[N][N];
int left[N], right[N];
int main()
{    
    char str[11];
    int t, n, m;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 1;i <= n;++i)
        {
            for (int j = 1;j <= m;++j)
            {
                scanf("%s", str);
                if (str[0] == 'R')
                    a[i][j] = 1;
                else
                    a[i][j] = 0;
                
                //得到每一行的高度
                if (a[i][j] == 0)
                    sum[i][j] = sum[i - 1][j] + 1;
                else
                    sum[i][j] = 0;
            }
        }
        int ans = 0;
        for (int i = 1;i <= n;++i)
        {
            left[1] = right[m] = 0;
            int l, r;
            for (int j = 2;j <= m;++j)
            {
                left[j] = 0;
                l = j - 1;
                while (l>=1 && sum[i][j] <= sum[i][l])
                {
                    left[j] += left[l] + 1;
                    l = l - left[l] - 1;
                }
            }
            for (int j = m - 1;j >= 1;--j)
            {
                right[j] = 0;
                r = j + 1;
                while (r <= n && sum[i][j] <= sum[i][r])
                {
                    right[j] += right[r] + 1;
                    r = r + right[r] + 1;
                }
            }
            for (int j = 1;j <= m;++j)
                ans = std::max(ans, (left[j] + right[j] + 1)*sum[i][j]);
        }
        printf("%d
", ans*3);
    }
    return 0;
}