深搜， 连通区域问题 cf 377A

Maze
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d
& %I64u
Submit Status

Description

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 Input

Input

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

Output

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

Input

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

Output

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

因为题目一定有解，所以可以找到一个连通区域，对这个连通区域进行深搜，找到最底下的节点，把连通区域看成是一棵树，从最小的树叶开始删除，则连通区域没有被破坏，所以一直从最考下的节点开始删除，删除到足够的个数时，返回，结束

#include<iostream>
#include<cmath>
#include<algorithm>
#include<stac
4000
k>
#include<cstring>
#include<cstdio>
using namespace std;

char maze[501][501];
int book[501][501];
int n,m,k;

void dfs(int r,int c)
{
if(r<0 || r>=n || c<0 || c>=m) return;
if(book[r][c]) return;
if(maze[r][c]!='.') return;
book[r][c]=1;
dfs(r,c-1); dfs(r-1,c); dfs(r,c+1); dfs(r+1,c);
if(k) maze[r][c]='X', k--;
}

int main()
{
while(cin>>n>>m>>k)
{
for(int i=0;i<n;i++)
scanf("%s",maze[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(maze[i]);
}

return 0;
}