Maze CF377A

题目

Pavel loves grid mazes. A grid maze is an n × m rectangle maze where each cell is either empty, or is a wall. You can go from one cell to another only if both cells are empty and have a common side.

Pavel drew a grid maze with all empty cells forming a connected area. That is, you can go from any empty cell to any other one. Pavel doesn't like it when his maze has too little walls. He wants to turn exactly k empty cells into walls so that all the remaining cells still formed a connected area. Help him.

Input

The first line contains three integers n, m, k (1 ≤ n, m ≤ 500, 0 ≤ k < s), where n and m are the maze's height and width, correspondingly, k is the number of walls Pavel wants to add and letter s represents the number of empty cells in the original maze.

Each of the next n lines contains m characters. They describe the original maze. If a character on a line equals ".", then the corresponding cell is empty and if the character equals "#", then the cell is a wall.

Output

Print n lines containing m characters each: the new maze that fits Pavel's requirements. Mark the empty cells that you transformed into walls as "X", the other cells must be left without changes (that is, "." and "#").

It is guaranteed that a solution exists. If there are multiple solutions you can output any of them.

Examples

Input

3 4 2
#..#
..#.
#...

Output

#.X#
X.#.
#...

Input

5 4 5
#...
#.#.
.#..
...#
.#.#

Output

#XXX
#X#.
X#..
...#
.#.#

题目大意

输入n,m,k表示迷宫的长宽和需要改变的数量

给定一个迷宫，.表示道路，# 表示墙壁。

题目给定的迷宫中由.构成的道路是连通的，现在让你改变k个.为X

当.变为X时，这个地方就像相当于墙壁

让你改变k个. 使迷宫中的.依然连通

算法：DFS

代码

#include<iostream>
#include<cstring>
using namespace std;
char a[550][550]; 
int b[550][550];
int d[4][2]={-1,0,1,0,0,-1,0,1};  
int m,n,k; 
void dfs(int x,int y)
{
	if(b[x][y] || a[x][y]!='.') return;
	b[x][y]=1; 
    for(int i=0;i<4;i++)
    {
    	int tx=x+d[i][0],ty=y+d[i][1];  
		if(tx<1 || tx>n || ty<1 || ty>m) continue; 
		dfs(tx,ty);
	}
    if(k) a[x][y]='X',k--;           
}
int main()
{
    while(cin>>n>>m>>k)
    {
        memset(b,0,sizeof(b));
		for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                cin>>a[i][j];    
        for(int i=1;i<=n && k;i++)
            for(int j=1;j<=m && k;j++)
            	dfs(i,j); 
        
        for(int i=1;i<=n;i++)
        {
        	for(int j=1;j<=m;j++)
                cout<<a[i][j]; 
            cout<<endl;
		}
                
    } 
    
    return 0;
}

关键点在DFS

每次从一个点dfs四个方向，不断地dfs直到最后的这个点无法再次dfs（也就是这个点是道路末端----改变为X不影响道路连通的点），此时将.变为X，然后k--，k控制改变数量。

一定要自己对着原来的迷宫试一试，你就明白dfs为什么最后改变为X了

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