City Game(最大子矩阵)
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.
InputThe first line of the input contains an integer K �C 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 �C reserved unit
F �C free unit
In the end of each area description there is a separating line.
OutputFor 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 // 题意:第一行测试组数T,然后 n ,m 代表矩阵大小 n行m列 ,只有 F 才能建,求最大子矩阵
有点复杂,但是,如果你做过 hdu1506 ,这题一下就想到了。
对于每行都建个 L[] , R[] 数组,
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define MX 1005
int n,m;
int h[MX][MX];
int L[MX][MX];
int R[MX][MX];
char mp[MX][MX]; void set_h()
{
memset(h,,sizeof(h));
for (int i=;i<=m;i++)
if (mp[][i]=='F')
h[][i]=;
for (int i=;i<=n;i++)
for (int j=;j<=m;j++)
if (mp[i][j]=='F')
h[i][j]=h[i-][j]+;
} void set_LR()
{
for (int i=;i<=n;i++)
{
for (int j=;j<=m;j++)
{
L[i][j]=j;
if (h[i][j]==) continue;
while (h[i][L[i][j]-]>=h[i][j])
L[i][j]=L[i][L[i][j]-];
}
}
for (int i=;i<=n;i++)
{
for (int j=m;j>=;j--)
{
R[i][j]=j;
if (h[i][j]==) continue;
while (h[i][R[i][j]+]>=h[i][j])
R[i][j]=R[i][R[i][j]+];
}
}
} int main()
{
int t;
cin>>t;
while (t--)
{
scanf("%d%d",&n,&m);
for (int i=;i<=n;i++)
{
for (int j=;j<=m;j++)
{
char sss[];
scanf("%s",sss);
mp[i][j]=sss[];
}
}
set_h();
set_LR();
int ans = ;
for (int i=;i<=n;i++)
{
for (int j=;j<=m;j++)
{
int area = (R[i][j]-L[i][j]+)*h[i][j];
if (area>ans) ans = area;
}
}
printf("%d\n",ans*);
}
return ;
}
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