Codeforces 377A

A. Maze

题目链接：http://codeforces.com/contest/377/problem/A

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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：
找k个空白快'.'填充成'X'，使剩下的空白快仍然组成联通块。
方法一：
dfs找到边缘块，从边缘块开始填充；如果你理解不了边缘块，可以看方法二。
代码1：

#include<iostream>
#include<cstdio>
using namespace std;
const int N=505;
int n,m,k;
int dir[4][2]={
    1,0,0,1,-1,0,0,-1
};
int vis[N][N]={
    0
};
char map[N][N];
void dfs(int x,int y) 
{
    for(int i=0;i<4;i++)
    {
        int a=x+dir[i][0];
        int b=y+dir[i][1];
        if(0<=a&&a<n&&0<=b&&b<m&&vis[a][b]==0&&map[a][b]=='.')
        {
            vis[a][b]=1;
            dfs(a,b);
        }
    }
    if(k)
    {
        k--;
        map[x][y]='X';
    }
}
int main()
{
    scanf("%d %d %d",&n,&m,&k);
    getchar();
    for(int i=0;i<n;i++)
    {
        gets(map[i]); 
    } 
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        if(vis[i][j]==0&&map[i][j]=='.')
        {
            vis[i][j]=1;
            dfs(i,j);
        }
    }
    for(int i=0;i<n;i++)
    puts(map[i]);
    return 0;
} 

 方法二：

逆向思维，先把所有的空白块'.'变成'X',记录个数为ans,再用dfs把ans-k个'X'变成'.'；这样就保证了所有的空白块都是连通块。

代码2：

#include<iostream>
#include<cstdio>
using namespace std;
const int N=505;
int n,m,k;
int cnt=0;
int ans=0;
int dir[4][2]={
    1,0,0,1,-1,0,0,-1
};
char map[N][N];
void dfs(int x,int y) 
{
    for(int i=0;i<4;i++)
    {
        int a=x+dir[i][0];
        int b=y+dir[i][1];
        if(0<=a&&a<n&&0<=b&&b<m&&map[a][b]=='X')
        {
            if(cnt==ans-k)return ;//控制'.'的个数为ans-k个 
            map[a][b]='.'; 
            cnt++;
            dfs(a,b);
        }
    }
}
int main()
{
    scanf("%d %d %d",&n,&m,&k);
    getchar();
    for(int i=0;i<n;i++)
    {
        gets(map[i]); 
    } 
    int x=0,y=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        if(map[i][j]=='.')
        {
             map[i][j]='X';
             ans++;
             x=i;
             y=j;
        }
    }
    map[x][y]='.';
    cnt++;
    dfs(x,y);
    for(int i=0;i<n;i++)
    puts(map[i]);
    return 0;
}