【CodeForces 525D】【dfs+思维】 Arthur and Walls 【只包含点和星的矩阵，需要将部分星变成点使点能组成矩形】_finally it is a day when arthur has enough money f-优快云博客

Arthur找到了一个理想的公寓，但房间布局不理想。本篇介绍一种算法，通过移除最少数量的墙壁，使每个房间形成矩形布局。

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描述：

Finally it is a day when Arthur has enough money for buying an apartment. He found a great option close to the center of the city with a nice price.

Plan of the apartment found by Arthur looks like a rectangle n × m consisting of squares of size 1 × 1. Each of those squares contains either a wall (such square is denoted by a symbol "*" on the plan) or a free space (such square is denoted on the plan by a symbol ".").

Room in an apartment is a maximal connected area consisting of free squares. Squares are considered adjacent if they share a common side.

The old Arthur dream is to live in an apartment where all rooms are rectangles. He asks you to calculate minimum number of walls you need to remove in order to achieve this goal. After removing a wall from a square it becomes a free square. While removing the walls it is possible that some rooms unite into a single one.

Input

The first line of the input contains two integers n, m (1 ≤ n, m ≤ 2000) denoting the size of the Arthur apartments.

Following n lines each contain m symbols — the plan of the apartment.

If the cell is denoted by a symbol "*" then it contains a wall.

If the cell is denoted by a symbol "." then it this cell is free from walls and also this cell is contained in some of the rooms.

Output

Output n rows each consisting of m symbols that show how the Arthur apartment plan should look like after deleting the minimum number of walls in order to make each room (maximum connected area free from walls) be a rectangle.

If there are several possible answers, output any of them.

   Examples
 

     input
   
5 5
.*.*.
*****
.*.*.
*****
.*.*.

     output
   
.*.*.
*****
.*.*.
*****
.*.*.

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

     output
   
***...*
..*...*
..*...*
..*...*
..*...*
*******

     input
   
4 5
.....
.....
..***
..*..

     output
   
.....
.....
.....
.....

题意：

给出一个n*m(1 ≤ n, m ≤ 2000)的表格，里面有'*'和'.'，求把最少的'*'移除掉，使得'.'所在的连通块是矩形。

思路：

2*2地考虑，如果2*2的格子里只有一个'*'，说明这个'*'要去掉，其他情况都不用去掉。然后去掉这个'*'后，又会对其他四个格子有影响。只要想到这个，就很好办了。
复杂度不太好估计。

推荐和CF 723D一起食用，题解见Here

代码：

#include <bits/stdc++.h>
#define pr(x) cout << #x << "= " << x << "  " ；
#define pl(x) cout << #x << "= " << x << endl;
#define ll __int64
using  namespace  std;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
const int N=2222;

int n,m;
char s[N][N];

void dfs(int x,int y){
  if(x<0 || x>=n-1 || y<0 || y>=m-1)return ;
  int sum=0,tx,ty;
  for(int i=0; i<2; i++){
    for(int j=0; j<2; j++){
      if(s[x+i][y+j]=='*'){
        sum++;
        tx=x+i;
        ty=y+j;
      }
    }
  }
  if(sum==1){//四个格子中有一个*就修改，并搜索因此影响到的区域
    s[tx][ty]='.';
    for(int i=-1; i<=1; i++)
      for(int j=-1; j<=1; j++)
        dfs(tx+i, ty+j);
  }
}

int  main(){
  #ifndef ONLINE_JUDGE
  freopen("in.txt","r",stdin);
  #endif

  read(n);read(m);
  for(int i=0; i<n; i++)scanf("%s", s[i]);
  for(int i=0; i<n-1; i++){//注意范围
    for(int j=0; j<m-1; j++){
      dfs(i, j);
    }
  }
  for(int i=0; i<n; i++)puts(s[i]);
  return 0;
}