HDU 1505 City Game （hdu1506 dp二维加强版）

F - City Game

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1505

Appoint description:

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

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int d[][];
int l[];
int r[];
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        int a,b;
        cin>>a>>b;
      //  memset(d,0,sizeof(d));
        for(int i=; i<a; i++)
        {
            for(int j=; j<b; j++)
            {
                char c[];
                cin>>c;
                if(c[]=='F')
                    d[i][j]=;
                else
                    d[i][j]=;
            }
        }
        /*
0 1 1 1 1 1                         0 1 1 1 1 1
1 1 1 1 1 1  (F=1,R=0,方便求和）    1 2 2 2 2 2
0 0 0 1 1 1  转化完就是右边矩阵     0 0 0 3 3 3
1 1 1 1 1 1                         1 1 1 4 4 4
1 1 1 1 1 1                         2 2 2 5 5 5
        */
        for(int i=; i<a; i++)
        {
            for(int j=; j<b; j++)
            {
                if(d[i][j]!=)
                    d[i][j]=d[i-][j]+;
            }
        }
     /*   printf("--------------->\n");
        for(int i=0; i<a; i++){
            for(int j=0; j<b; j++)
                printf("%d ",d[i][j]);
            printf("\n");
        }

           printf("----------------->\n");*/
        int max=;
        for(int i=; i<a; i++){
            for(int j=; j<b; j++){
                l[j]=j;
                while(l[j]>&&d[i][l[j]-]>=d[i][j])
                       l[j]=l[l[j]-];
            }
            for(int j=b-; j>-; j--){
                r[j]=j;
                while(r[j]<b-&&d[i][r[j]+]>=d[i][j])
                    r[j]=r[r[j]+];
            }
            for(int j=; j<b; j++)
                if(max<((r[j]-l[j]+)*d[i][j]))
                    max=((r[j]-l[j]+)*d[i][j]);
        }
        cout<<max*<<endl;
    }
    return ;
}