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