HDU1505_City Game【最大完全子矩阵】

City Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4727    Accepted Submission(s): 2013

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
 
Source

Southeastern Europe 2004s

题目大意：给你一个M*N的区域。R代表被占用，F代表空闲。每块空闲

的区域价值3美金，求全部由空闲区域围成的矩形的价值

思路：1506的升级版。不同的是，1505是二维的。把2维转换为以每一

行为底，组成的最大面积就转换成了1506题。

参考1506题解题报告：http://blog.csdn.net/lianai911/article/details/40208265

其中h[i][j]表示第i行为底，第j列上方连续空闲位置的高度。

遍历计算出该点向左右两边延伸的左右边界，从而计算出面积，最终比较

计算出最大面积。乘以3就是最终的价值。

#include
#include

int h[1100][1100],l[1100],r[1100];
int main()
{
    int K,M,N;
    char ch[4];
    scanf("%d",&K);
    while(K--)
    {
        memset(h,0,sizeof(h));
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        scanf("%d%d",&M,&N);
        for(int i = 1; i <= M; i++)
        {
            for(int j = 1; j <= N; j++)
            {
                scanf("%s",ch);
                if(ch[0] == 'F')
                    h[i][j] = h[i-1][j] + 1;
                else
                    h[i][j] = 0;
            }
        }

        __int64 MaxArea = -0xffffff0;

        for(int i = 1; i <= M; i++)
        {
            for(int j = 1; j <= N; j++)
                l[j] = r[j] = j;
            l[0] = 1;
            r[N+1] = N;
            h[i][0] = -1;
            h[i][N+1] = -1;
            for(int j = 1; j <= N; j++)
            {
                while(h[i][l[j]-1] >= h[i][j])
                    l[j] = l[l[j]-1];
            }
            for(int j = N; j >= 1; j--)
            {
                while(h[i][r[j]+1] >= h[i][j])
                    r[j] = r[r[j]+1];
            }
            for(int j = 1; j <= N; j++)
            {
                if(h[i][j]*(r[j]-l[j]+1) > MaxArea)
                    MaxArea = h[i][j]*(r[j]-l[j]+1);
            }
        }
        printf("%I64d\n",MaxArea*3);
    }


    return 0;
}

/*
输入错误。
for(int i = 1; i <= M; i++)
{
    for(int j = 1; j <= N; j++)
    {
        getchar();//吸收空格和换行符
        char ch = getchar();
        if(ch == 'F')
            h[i][j] = h[i-1][j] + 1;
        else
            h[i][j] = 0;
    }
}
*/