Codeforces #423-Div. 2-B. Black Square

B. Black Square

time limit per test1 second
memory limit per test256 megabytes

题目：

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

input5 4WWWWWWWBWWWBWWBBWWWWoutput5input1 2BBoutput-1input3 3WWWWWWWWWoutput1

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：

题意很简单，就是说B代表黑色，问最少涂黑几个黑色可以让所有的黑色方块组成正方形（只能把白色的变黑不能把黑色变白），无解输出-1

思路：

我的思路很暴力，找出所有黑色方块中四个方向最远的四个方向，算他们的能组成的最长边，然后如果最长边比输出矩形的宽要长就是无解，否则输出，（最长边的平方-B的数量）

代码：

#include <bits/stdc++.h>using namespace std;int n,m,flag=1,b,w;char mp[105][105];int bian(){    int maxx=0,minx=INFINITY,miny=INFINITY,maxy=0;    for(int i=0;i<n;i++)    {        for(int j=0;j<m;j++)        {            if(mp[i][j]=='B')            {                if(maxx<i)maxx=i;                if(minx>i)minx=i;                if(maxy<j)maxy=j;                if(miny>j)miny=j;            }        }    }    int l=maxx-minx+1,h=maxy-miny+1;    if(max(l,h)<=min(n,m))    {        return max(l,h)*max(l,h)-b;    }    else    {        return -1;    }}int main(){    cin>>n>>m;getchar();    for(int i=0;i<n;i++)    {        for(int j=0;j<m;j++)        {            scanf("%c",&mp[i][j]);            if(mp[i][j]=='B')            {                if(flag)flag=0;                b++;            }            else            {                w++;            }        }        getchar();    }    if(flag)printf("1\n");    else    {        printf("%d\n",bian());    }    return 0;}