HDU1505-City Game（记忆化搜索）

City Game

http://acm.hdu.edu.cn/showproblem.php?pid=1505

Problem Description

Bob is a strategy game programming specialist. In his new city building game the gaming environment is as follows: a city is built up by areas, in which there are streets, trees,factories and buildings. There is still some space
in the area that is unoccupied. The strategic task of his game is to win as much rent money from these free spaces. To win rent money you must erect buildings, that can only be rectangular, as long and wide as you can. Bob is trying to find a way to build
the biggest possible building in each area. But he comes across some problems – he is not allowed to destroy already existing buildings, trees, factories and streets in the area he is building in.

Each area has its width and length. The area is divided into a grid of equal square units.The rent paid for each unit on which you're building stands is 3$.

Your task is to help Bob solve this problem. The whole city is divided into K areas. Each one of the areas is rectangular and has a different grid size with its own length M and width N.The existing occupied units are marked with the symbol R. The unoccupied
units are marked with the symbol F.

 

Input

The first line of the input contains an integer K – determining the number of datasets. Next lines contain the area descriptions. One description is defined in the following way: The first line contains two integers-area length M<=1000
and width N<=1000, separated by a blank space. The next M lines contain N symbols that mark the reserved or free grid units,separated by a blank space. The symbols used are:

R – reserved unit

F – free unit

In the end of each area description there is a separating line.

 

Output

For each data set in the input print on a separate line, on the standard output, the integer that represents the profit obtained by erecting the largest building in the area encoded by the data set.

 

Sample Input

2
5 6
R F F F F F
F F F F F F
R R R F F F
F F F F F F
F F F F F F

5 5
R R R R R
R R R R R
R R R R R
R R R R R
R R R R R

 

Sample Output

45
0

解题心得：

1、这个题是POJ2559-Largest
Rectangle in a Histogram的升级版，关于左右状态的转移就不说了，前面那个链接里面说了。就说说怎么把这个题和前面那个题联系起来解决。

2、这个题有一个处理技巧就是将同一列里面连续的'F'个数加起来，就变成了列宽，在一个’F‘的左方列宽大于'F'的才能够进行left的状态转移，右方同理，这样得到的有left和right还有列宽就可以直接得到面积了，这个技巧的启示来自于这个题：矩阵求最大和。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1100;
char maps[maxn][maxn];
int h[maxn][maxn];
int r[maxn],l[maxn];
int n,m;
int Max = 0;

void get_leftEdge(int x)
{
    for(int j=2; j<=m; j++)
    {
        if(h[x][j-1] >= h[x][j])
            l[j] = l[j-1];
    }
}

void get_rightEdge(int x)
{
    for(int j=m-1; j>=1; j--)
    {
        if(h[x][j+1] >= h[x][j])
            r[j] = r[j+1];
    }

}

void get_ans(int x)
{
    for(int j=1; j<=m; j++)
        if(Max < h[x][j] * (r[j] - l[j] + 1))
            Max = h[x][j] * (r[j] - l[j] + 1);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        Max = 0;
        memset(h,0,sizeof(h));
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                scanf("%s",&maps[i][j]);
                if(maps[i][j] == 'F')
                {
                    h[i][j] = h[i-1][j] + 1;
                }
                l[j] = r[j] = j;
            }
            get_leftEdge(i);//处理maps[i][j]在i行左方的边界
            get_rightEdge(i);//右方边界
            get_ans(i);
        }
        printf("%d\n",Max*3);
    }
}