Pavel loves grid mazes（CodeForce 377A）

题目如下：

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#..
...#
.#.#

哈哈！机智的你应该已经知道题意了吧、、、

我们来稍微理一下题意，给你一个nxm的矩阵，#代表墙，·代表可以通过。然后再告诉你一个K值。

问你，将k个点变为X，使得剩下的点依然是连通的。让你输出这个图、（之前给出的图是保证点是连通的）。

基本思路：

因为原图是连通的，所以意思就是对这个图进行DFS，肯定可以使所有的点都在它的树上。

既然是这样，那我们就依次从树的最下端开始倒着删除节点，一直到删除了K个为止。

因为我们是从最下端开始删除的，所以能够保证上面的节点肯定是互相连通的。

所以，最后把删除掉的点改成X，直接输出矩阵就可以了。

代码如下：

本人大水比一枚、代码如有问题，请积极指正、、

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#include #include #include #include using namespace std; char mp[520][520]; bool vis[520][520]; int f[4][2]={0,1,0,-1,1,0,-1,0}; int n,m,k; void init() { memset(vis,false,sizeof(vis)); } 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; for(int i=0;i<4;i++) DFS(x+f[i][0],y+f[i][1]); if(k) mp[x][y]='X',k--; } int main() { while(~scanf("%d%d%d",&n,&m,&k)) { init(); for(int i=0;i