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$.
Your task is to help Bob solve this problem. The whole city is divided into K areas. Each one of the areas is rectangular and has a different grid size with its own length M and width N.The existing occupied units are marked with the symbol R. The unoccupied units are marked with the symbol F.
Each area has its width and length. The area is divided into a grid of equal square units.The rent paid for each unit on which you're building stands is 3$.
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.
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
1506的升级版。换成在二维上的了=。=
依旧是那个方法=。=复杂的方法
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=1100;
char mp[maxn][maxn];
int a[maxn][maxn];
int l[maxn],r[maxn];
int n,m,t;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
getchar();
for(int i=1;i<=m;i++)
a[0][i]=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
cin>>mp[i][j];
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
if(mp[j][i]=='F')
a[j][i]=a[j-1][i]+1;
else
a[j][i]=0;
}
}
int ans=-1;
for(int i=1;i<=n;i++)
{
l[1]=1;
r[m]=m;
for(int j=2;j<=m;j++)
{
if(a[i][j]==0)
continue;
int t=j;
while(t>1&&a[i][j]<=a[i][t-1])
t=l[t-1];
l[j]=t;
}
for(int j=m-1;j>=1;j--)
{
if(a[i][j]==0)
continue;
int t=j;
while(t<m&&a[i][j]<=a[i][t+1])
t=r[t+1];
r[j]=t;
}
for(int j=1;j<=m;j++)
{
if((r[j]-l[j]+1)*a[i][j]>ans)
ans=(r[j]-l[j]+1)*a[i][j];
}
}
printf("%d\n",ans*3);
}
return 0;
}
简单的方法=。=，设一个num数组。耗的内存和时间都少了（事实上时间差点儿相同）
可是理解起来去easy些=。=
附张图：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=1100;
char mp[maxn][maxn];
int num[maxn];
int n,m,t;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
getchar();
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
cin>>mp[i][j];
}
int ans=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(mp[i][j]=='F')//列中连续个数
num[j]++;
else
num[j]=0;
}
for(int j=1;j<=m;j++)
{
if(!num[j])
continue;
int cnt=1;
for(int k=1;j-k>=1&&num[j]<=num[j-k];k++)
++cnt;
for(int k=1;j+k<=m&&num[j]<=num[j+k];k++)
++cnt;
if(ans<num[j]*cnt)
ans=num[j]*cnt;
}
}
printf("%d\n",ans*3);
}
return 0;
}
附张图：
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5059286.html，如需转载请自行联系原作者
