HDU 之 City Game

City Game

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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 file 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 file 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

算法解析：
 扫描线模拟！

R F F F F F                 0 1 2 3 4 5
F F F F F F                 1 2 3 4 5 6
R R R F F F      ====》》》》0 0 0 1 2 3
F F F F F F                 1 2 3 4 5 6
F F F F F F                 1 2 3 4 5 6

R则为0， 连续的F则赋 1++ ；（这是有用的，其实可以判断连续的宽度）

#include <stdio.h>
#include <string.h>

int map[1001][1001];

int main()
{
	int t;
	scanf("%d", &t );
	int i, j, k;
    int n, m;  
	char ch;
	while(t--)
    {
		scanf("%d %d%*c", &n, &m );
		memset(map, 0, sizeof(0)); //每次都得初始化
		for(i=1; i<=n; i++)
		{
			for(j=1; j<=m; j++)
			{
				ch=getchar();
				while(ch!='F' && ch!='R' )
				{
					ch=getchar();
				}
				if(ch=='F')
				{
					map[i][j] = map[i][j-1] + 1;
				}
				else
				{
					map[i][j] = 0;
				}
			}
		} //map的创建,注意对字符与空格的处理,然后转化为数字存进map
		  //存数方式要注意
		int min; 
		int area=0; //初始面积为0
		int high; //定义高度
		for(i=1; i<=n; i++)
		{
			for(j=1; j<=m; j++)
			{
				if(map[i][j]!=0) //访问的数据>0 : (F)
				{
					min=map[i][j] ;
					if( area < map[i][j] )
					{
						area = map[i][j] ;
					}
					high =1; //初始高度自身
					for(k=i-1; k>=1; k--)
					{
						if(map[k][j]==0 )
							break;
						else
						{
							high++; //高度增加
							if(map[k][j] < min )
							{
								min = map[k][j] ;
							}
							if(area < high*min )
							{
								area = high*min;
							}
						}
					}
				}
			}
		}
		printf("%d
", area*3 );
	}
	return 0;
}