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
Copy
5 4 WWWW WWWB WWWB WWBB WWWW
Output
Copy
5
Input
Copy
1 2 BB
Output
Copy
-1
Input
Copy
3 3 WWW WWW WWW
Output
Copy
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.
题意 :
给你一个广场，上面有黑色和白色的瓷砖，你可以把白色的瓷砖换成黑色的瓷砖，求最少要多少块黑色瓷砖来换才能构成一个黑色的正方形。
思路:
找出两个黑色瓷砖的最远的距离即正方形的边长，如果边长大于原本广场的最小的边输出-1，要不然输出这个正方形所需的瓷砖总数-已有黑色瓷砖数。
#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
char a[101][101];
int m,n;
cin>>m>>n;
int sum=0;
int maxx=0,maxy=0,minx=101,miny=101;
for(int i=0; i<m; i++)
for(int j=0; j<n; j++)
cin>>a[i][j];
for(int i=0; i<m; i++)
for(int j=0; j<n; j++)
{
if(a[i][j]=='B')
{
sum++;
if(i>maxx)
maxx=i;
if(i<minx)
minx=i;
if(j<miny)
miny=j;
if(j>maxy)
maxy=j;
}
}
int ans=max(maxx-minx,maxy-miny);
if(ans+1>m||ans+1>n)
printf("-1");
else if(sum==0)
printf("1");
else
printf("%d",(ans+1)*(ans+1)-sum);
return 0;
}