A group of K friends is going to see a movie. However, they are too late to get good tickets, so they are looking for a good way to sit all nearby. Since they are all science students, they decided to come up with an optimization problem instead of going on with informal arguments to decide which tickets to buy.
The movie theater has R rows of C seats each, and they can see a map with the currently available seats marked. They decided that seating close to each other is all that matters, even if that means seating in the front row where the screen is so big it’s impossible to see it all at once. In order to have a formal criteria, they thought they would buy seats in order to minimize the extension of their group.
The extension is defined as the area of the smallest rectangle with sides parallel to the seats that contains all bought seats. The area of a rectangle is the number of seats contained in it.
They’ve taken out a laptop and pointed at you to help them find those desired seats.
Input
Each test case will consist on several lines. The first line will contain three positive integers R, C and K as explained above (1 <= R,C <= 300, 1 <= K <= R × C). The next R lines will contain exactly C characters each. The j-th character of the i-th line will be ‘X’ if the j-th seat on the i-th row is taken or ‘.’ if it is available. There will always be at least K available seats in total.
Input is terminated with R = C = K = 0.
Output
For each test case, output a single line containing the minimum extension the group can have.
Sample Input：
3 5 5
…XX
.X.XX
XX…
5 6 6
..X.X.
.XXX..
.XX.X.
.XXX.X
.XX.XX
0 0 0
Sample Output：
6
9
题意：求一个“.”的个数大于等于k的最小矩形。显然暴力会超时，当时有想到求二维前缀和（自己造的词），但是没有想到好的方法实现。看了题解后，才知道可以在输入的时候就求前缀和（指以该点为右下角的矩形中“.”的个数）。实现方法见代码。之后，固定行的上限与下限，用尺取法枚举列。算每个小矩形时要注意减一。比如一个小矩形左上角的点为（x1,y1）,右下角的点为（x2,y2）。那么该小矩形中的个数为sum[x2][y2]-sum[x2][y1-1]-sum[x1-1][y2]+sum[x1-1][y1-1]。
PS:注意因为有i-1，j-1,所以数组下标都从1开始。
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
using namespace std;
int main()
{
char map[310][310];
int sum[310][310];
int r,c,k;
while(scanf("%d%d%d",&r,&c,&k)!=EOF)
{
if(r==0&&c==0&&k==0)
break;
int i,j;
memset(sum,0,sizeof(sum));
char ch;
for(i=1;i<=r;i++)
{
for(j=1;j<=c;j++)
{
cin>>ch;
int t;
t=(ch=='.');
sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+t;
}
}
int ans=r*c;
for(i=1;i<=r;i++) //i,j分别为行的上限和下限
for(j=i;j<=r;j++)
{
int l,rr;
l=rr=1; //尺取找符合条件的列的上限与下限
while(1)
{
//这里特别注意计算小矩形内“.”的个数的方法，这里的（i,j）代表的不是一个点，而是一个小格
while(sum[j][rr]-sum[j][l-1]-sum[i-1][rr]+sum[i-1][l-1]<k&&rr<=c)
rr++;
if(sum[j][rr]-sum[j][l-1]-sum[i-1][rr]+sum[i-1][l-1]<k)
break;
if(sum[j][rr]-sum[j][l-1]-sum[i-1][rr]+sum[i-1][l-1]>=k)
{
ans=min(ans,(rr-l+1)*(j-i+1));
}
l++;
}
}
cout<<ans<<endl;
}
return 0;
}