hdu 1937 Finding Seats

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 889    Accepted Submission(s): 290

Problem Description

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

 

Source

ACM South American Programming Contest 2007

 

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题目是要求最小的矩形面积，要求矩形区域内有足够k个人坐的座位，如此实际上就是求矩形的有效面积，先求二维前缀和，然后进行暴力，结果超时，毕竟四层循环，想到尺取，可是怎么移动横纵坐标，都移动？不可，会有遗漏的情况，可以只对一个坐标尺取，这里是保持行的范围不变，对列进行尺取。

代码：

#include <iostream>
#include <cstdio>
#define MAX 300
using namespace std;
int r,c,k;
int sum[MAX + 1][MAX + 1];
inline int get(int l1,int l2,int r1,int r2) {
    return sum[l2][r2] - sum[l1 - 1][r2] - sum[l2][r1 - 1] + sum[l1 - 1][r1 - 1];
}
int main() {
    while(~scanf("%d%d%d",&r,&c,&k) && r && c && k) {
        int ans = r * c + 1;
        for(int i = 1;i <= r;i ++) {
            getchar();
            for(int j = 1;j <= c;j ++) {
                char ch = getchar();
                if(ch == '.') sum[i][j] = 1;
                else sum[i][j] = 0;
                sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
            }
        }
        for(int i = 1;i <= r;i ++) {
            for(int j = 1;j <= r;j ++) {
                int left = 1,right = 1;
                while(right <= c && left <= right) {
                    if(get(i,j,left,right) < k) right ++;
                    else {
                        ans = min(ans,(j - i + 1) * (right - left + 1));
                        left ++;
                    }
                }
            }
        }
        printf("%d
",ans);
    }
}