HDU1505_City Game【最大完全サブマトリクス】

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
M*Nのエリアをあげます.Rは占有を表し、Fは空きを表す.各ブロックの空き
の区域の価値は3ドルで、すべて暇な区域から囲む矩形の価値を求めます
構想:1506のアップグレード版.違いは1505が2次元です.2次元を1つずつに変換
行動の底で,組成の最大面積は1506題に変換された.
参考1506問題解決報告書:http://blog.csdn.net/lianai911/article/details/40208265
ここで、h[i][j]は、i番目の動作の底、j番目の列の上の連続した空き位置の高さを表す.
この点が左右に延びる左右の境界を遍歴して算出し、面積を算出し、最終的に比較する
最大面積を算出します.3を掛けることが最終的な価値です.

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

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
",MaxArea*3);
    }


    return 0;
}

/*
    。
for(int i = 1; i <= M; i++)
{
    for(int j = 1; j <= N; j++)
    {
        getchar();//        
        char ch = getchar();
        if(ch == 'F')
            h[i][j] = h[i-1][j] + 1;
        else
            h[i][j] = 0;
    }
}
*/