Maze

最新推荐文章于 2022-11-14 11:06:14 发布

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#dfs

本文介绍了一个迷宫改造问题，目标是在不破坏连通性的前提下，将指定数量的空地转化为墙壁。通过深度优先搜索（DFS）算法实现这一目标，并详细展示了如何确保迷宫在改造后依然保持连通。

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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.

Example
Input

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

Output

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

Input

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

Output

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

#include<stdio.h>
#include<cstring>
#include<queue>
using namespace std;
const int Max=501;

//这个题感觉还是蛮简单的，主要是一开始理解错了题意，以为他要好几个连通区间
//就只有一个入口
char Map[Max][Max];
int vis[Max][Max];
int n,m,k;

int check(int x,int y)
{
    if(vis[x][y]||x<1||y<1||x>n||y>m||Map[x][y]!='.')
        return 1;
    return 0;
}

void dfs(int x,int y)
{
    if(check(x,y))
        return;
    vis[x][y]=1;
    dfs(x,y+1);
    dfs(x,y-1);
    dfs(x-1,y);
    dfs(x+1,y);
    if(k)
        Map[x][y]='X',k--;//撤出一个点的时候，将这个点改变
    //一开始用的while(),想在明白了，我们每一个函数都是对一个点进行的操作
   //还是加深了自己对递归的理解
}

int main()
{
   memset(Map,0,sizeof(Map));
   memset(vis,0,sizeof(vis));
   scanf("%d %d %d",&n,&m,&k);
   for(int i=1;i<=n;i++)
       scanf("%s",Map[i]+1);
   for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
        if(Map[i][j]=='.')
            dfs(i,j);

   for(int i=1;i<=n;i++)
      printf("%s\n",Map[i]+1);
   return 0;
}