Codeforces Round #423 (Div. 2) B. Black Square(思路)

题目：

B. Black Square

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp has a checkered sheet of paper of size n × m. Polycarp painted some of cells with black, the others remained white. Inspired by Malevich's "Black Square", Polycarp wants to paint minimum possible number of white cells with black so that all black cells form a square.

You are to determine the minimum possible number of cells needed to be painted black so that the black cells form a black square with sides parallel to the painting's sides. All the cells that do not belong to the square should be white. The square's side should have positive length.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the sizes of the sheet.

The next n lines contain m letters 'B' or 'W' each — the description of initial cells' colors. If a letter is 'B', then the corresponding cell is painted black, otherwise it is painted white.

Output

Print the minimum number of cells needed to be painted black so that the black cells form a black square with sides parallel to the painting's sides. All the cells that do not belong to the square should be white. If it is impossible, print -1.

Examples

input

5 4WWWWWWWBWWWBWWBBWWWW

output

5

input

1 2BB

output

-1

input

3 3WWWWWWWWW

output

1

Note

In the first example it is needed to paint 5 cells — (2, 2), (2, 3), (3, 2), (3, 3) and (4, 2). Then there will be a square with side equal to three, and the upper left corner in (2, 2).

In the second example all the cells are painted black and form a rectangle, so it's impossible to get a square.

In the third example all cells are colored white, so it's sufficient to color any cell black.

思路：

给了一张n*m的图，W代表白色，B代表黑色，问的是在这张图中至少添加几个黑色可以在这张图中使黑色连成一个正方形。如果不能连成就输出-1

遍历这张图，然后找出横纵坐标的极值，然后算出来最大的差+1就是正方形的边长，我们只需要判断边长是不是越界，然后输出结果就好

代码：

#include <cstdio>#include <cstring>#include <string>#include <set>#include <iostream>#include <cmath>#include <stack>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define debug() puts("what the fuck!!!")#define N (200000+50)#define ll long longusing namespace std;char map[110][110];int main(){int n,m,tot=0;int x1=inf,x2=0,y1=inf,y2=0;scanf("%d%d",&n,&m);for(int i=0; i<n; i++){scanf("%s",map[i]);for(int j=0; j<m; j++){if(map[i][j]=='B'){tot++;x1=min(x1,i);x2=max(x2,i);y1=min(y1,j);y2=max(y2,j);}}}if(!tot){puts("1");return 0;}int c=max(x2-x1,y2-y1)+1;if(c>n||c>m)puts("-1");elseprintf("%d\n",c*c-tot);return 0;}