CodeForces 377 A. Maze

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本文介绍了一种使用深度优先搜索（DFS）算法解决迷宫问题的方法，目标是在给定的迷宫中恰当地添加指定数量的墙壁，以保持所有空地相连。通过DFS遍历整个迷宫，并在满足条件的情况下将空地变为墙壁。

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DFS找末端的空地变成墙。。。。。。

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.

Sample test(s)

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

output
#.X#
X.#.
#...

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

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

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char mp[512][512];
bool vis[512][512];

int n,m,k;

void dfs(int x,int y)
{
    if(x<0||x>=n||y<0||y>=m) return ;
    if(mp[x][y]!='.') return ;
    if(vis[x][y]) return ;
    vis[x][y]=true;
    dfs(x+1,y); dfs(x-1,y);
    dfs(x,y+1); dfs(x,y-1);
    if(k) mp[x][y]='X',k--;
}

int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<n;i++) scanf("%s",mp[i]);
    for(int i=0;i<n&&k;i++)
    {
        for(int j=0;j<m&&k;j++)
        {
            dfs(i,j);
        }
    }
    for(int i=0;i<n;i++) puts(mp[i]);
    return 0;
}