Finding Seats--尺取法

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博客围绕电影座位安排问题展开，有r排c列座位，部分已被占，k个人观影想坐得尽量集中。需找出包含至少k个空座位的最小矩形。解题采用尺取法，用数组记录特定矩形内空座位数，通过枚举列与行的起点和终点求解。

题目

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

有r排c列座位，其中有一些是已经有人占的（‘X’），其余的是没人占的（‘.’）。一共有k个人去看电影，他们想挨的越近越好，也就是越不分散越好，求一个最小的矩形，使得包含的空座位数大于等于k。

分析

这个题要用尺取法。首先用一个num[ i ][ j ]数组，表示左上角为（1,1），右下角为（i，j）的矩形中空座位的个数，然后，枚举列的起点和终点（i，j），然后在枚举行的起点和终点（p，t）的时候用尺取法，求出最小的矩形。

代码

#include<iostream>
#include<cstdio>
#include<string.h>
#include<stack>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f
int r,c,k;
char mp[400][400];
int num[400][400];
int main()
{
	while(~scanf("%d%d%d",&r,&c,&k))
	{
		if(r==0&&c==0&&k==0)
			break;
		for(int i=1;i<=r;i++)
			scanf("%s",mp[i]+1);
		memset(num,0,sizeof(num));
		for(int i=1;i<=r;i++)
		{
			int tmp=0;
			for(int j=1;j<=c;j++)
			{
				if(mp[i][j]=='.')
					tmp++;
				num[i][j]=num[i-1][j]+tmp;
			}
		}
		int ans=1e9;
		for(int i=1;i<=c;i++)//列的起点 
		{
			for(int j=i;j<=c;j++)//列的终点 
			{
				int p=1;//行起点 
				for(int t=1;t<=r;t++)//行终点 
				{
					while(num[t][j]-num[t][i-1]-num[p-1][j]+num[p-1][i-1]>=k)
					{
						ans=min(ans,(j-i+1)*(t-p+1));
						p++;
					}
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}