Codeforces 525D Arthur and Walls 【Dfs+思维】_d. arthur and walls-优快云博客

本文介绍了一种算法，旨在通过最小化更改数量将地图上的障碍物变为通路，确保所有可通行区域形成矩形房间。通过深度优先搜索(DFS)识别必须转换的障碍物。

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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
.....
.....
.....
.....

题目大意：

让你将星号改变成点，使得联通的点都是矩形，尽量将更少的星号进行改变。

思路：

1、介个题思路走对了，毛线，一直莫名re/TLE在92，实在是改不动代码了，参考了巨巨的代码~

代码参考自：http://blog.youkuaiyun.com/kalilili/article/details/44724351

2、首先通过观察可知，对应一个星号如果周围有三个点，并且这三个点是呈拐弯形式出现的点，那么这个星号一定是需要被改变成点的。e.g：

*	.
.	.

能够观察出来，此时点的联通形状此时有拐弯的情况发生，那么这个星号一定要变成点才行。

3、那么我们将一些星号改变成点之后，会影响其他的星号也能变成点，那么此时我们Dfs处理即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
char a[2010][2012];
int fx[5]={0,1,0,-1,0};
int fy[5]={1,0,-1,0,1};
int n,m;
void Dfs(int x,int y)
{
    if(a[x][y]=='*')
    {
        for(int i=0;i<4;i++)
        {
           int x1=x+fx[i];
            int y1=y+fy[i];
            int x2=x+fx[i+1];
            int y2=y+fy[i+1];
            int x3,y3;
            if(fx[i]!=0)x3=x+fx[i];
            else x3=x+fx[i+1];
            if(fy[i]!=0)y3=y+fy[i];
            else y3=y+fy[i+1];
            if(a[x1][y1]=='.'&&a[x2][y2]=='.'&&a[x3][y3]=='.')
            {
                a[x][y]='.';
                for(int xx=x-1;xx<=x+1;xx++)
                {
                    for(int yy=y-1;yy<=y+1;yy++)
                    {
                        if(xx>=0&&xx<n&&yy>=0&&yy<m)
                        Dfs(xx,yy);
                    }
                }
                return ;
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='*')
                Dfs(i,j);
            }
        }
        for(int i=0;i<n;i++)
        {
            printf("%s\n",a[i]);
        }
    }
}