zoukankan      html  css  js  c++  java
  • B

    Problem description

    A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with n rows and m columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.

    Input

    The first line of the input data contains numbers n and m (1 ≤ n, m ≤ 50), n — amount of lines, and m — amount of columns on Bob's sheet. The following n lines contain m characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.

    Output

    Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.

    Examples

    Input

    6 7
    .......
    ..***..
    ..*....
    ..***..
    ..*....
    ..***..

    Output

    ***
    *..
    ***
    *..
    ***

    Input

    3 3
    ***
    *.*
    ***

    Output

    ***
    *.*
    ***
    解题思路:输出最小覆盖所有星号的矩形。
    AC代码:
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 int main(){
     4     int n,m,rou,rod,colt,cort;char s[55][55];
     5     cin>>n>>m;getchar();
     6     rou=n-1,rod=0,colt=m-1,cort=0;
     7     for(int i=0;i<n;++i)
     8         for(int j=0;j<m;++j)
     9             cin>>s[i][j];
    10     for(int i=0;i<n;++i){
    11         int j=0,k=m-1;
    12         while(j<m && s[i][j]=='.')j++;
    13         while(k>=0 && s[i][k]=='.')k--;
    14         if(j<=k){
    15             colt=min(colt,j);cort=max(cort,k);//确定列的左右最大范围
    16             rou=min(i,rou);rod=max(rod,i);//确定行的上下最大范围
    17         }
    18     }
    19     for(int i=rou;i<=rod;++i){
    20         for(int j=colt;j<=cort;++j)cout<<s[i][j];
    21         cout<<endl;
    22     }
    23     return 0;
    24 }
    
    
  • 相关阅读:
    github使用
    部署flask
    docker部署路飞学城
    centos7安装dnsmasq局域网dns
    消息队列rabbitmq
    记录腾讯云中矿机病毒处理过程(重装系统了fu*k)
    Golang基础
    git协同开发
    gitlab与pycharm结合
    github与gitlab与git三个基佬的故事
  • 原文地址:https://www.cnblogs.com/acgoto/p/9136644.html
Copyright © 2011-2022 走看看