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.
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.
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
#include<stdio.h> #include<string.h> int map[400][400]; char str[400][400]; int tmin,k; int calculate(int x1,int y1,int x2,int y2){ int point=map[x1][y1]-map[x2-1][y1]-map[x1][y2-1]+map[x2-1][y2-1];//判断这一范围内的顶点数 int area=(x1-x2+1)*(y1-y2+1);//判断这一范围内的面积 if(point>=k&&area<tmin)//如果顶点数足够并且面积可以缩小，则更新最小值 tmin=area; return point; } int main(){ int x,y; while(scanf("%d%d%d",&x,&y,&k)!=EOF){ if(x==0&&y==0&&k==0) break; memset(str,0,sizeof(str)); memset(map,0,sizeof(map)); getchar(); for(int i=1;i<=x;i++){ scanf("%s",str[i]+1); getchar(); int sum=0; for(int j=1;j<=y;j++){ if(str[i][j]=='.') sum++; map[i][j]=map[i-1][j]+sum;//构建map数组，类似于动态规划的思想 } } tmin=1000000; for(int x1=x;x1>=1;x1--){//从下到上 if(map[x1][y]<k) break; for(int x2=1;x2<=x1;x2++){//从上到下 if(map[x1][y]-map[x2-1][y]<k) break; int y1=1,y2=1; while(y1<=y&&y2<=y){//两个顶点从左到右 if(calculate(x1,y1,x2,y2)>=k) y2++; else{ if(y1==y) break; y1++; } } } } printf("%d ",tmin); } return 0; }