zoukankan      html  css  js  c++  java
  • Round #423 B. Black Square(Div.2)

    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 4
    WWWW
    WWWB
    WWWB
    WWBB
    WWWW
    output
    5
    input
    1 2
    BB
    output
    -1
    input
    3 3
    WWW
    WWW
    WWW
    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.

     1 #include <iostream>
     2 #include <stdio.h>
     3 using namespace std;
     4 int main(){
     5     int n,m,ans=0;
     6     scanf("%d%d",&n,&m);
     7     char a;
     8     int imi=105,jmi=105,ima=0,jma=0;
     9     for(int i=1;i<=n;i++)
    10     for(int j=1;j<=m;j++){
    11             cin>>a;   //scanf("%c",&a);
    12             if(a=='B'){
    13                 ans++;
    14                 imi=min(imi,i);
    15                 ima=max(ima,i);
    16                 jmi=min(jmi,j);
    17                 jma=max(jma,j);
    18             }
    19     }
    20     int k=max(ima-imi,jma-jmi)+1;
    21     if(ans==0)  printf("1
    ");
    22     else if(n>=k&&m>=k) printf("%d
    ",k*k-ans);
    23     else printf("-1
    ");
    24     return 0;
    25 }
  • 相关阅读:
    源码实现 --> strcmp
    源码实现 --> strdel
    源码实现 --> strcpy
    SoC的Testbench中的简易bus_monitor(加入print函数)
    debian 安装后需做的几件事
    使用Perl合并文件
    一个简单的Verilog计数器模型
    Environment Modules简单使用
    Git push “fatal: Authentication failed ”
    使用SystemC进行硬件仿真
  • 原文地址:https://www.cnblogs.com/z-712/p/7307883.html
Copyright © 2011-2022 走看看