【Codeforces - 378C】Maze（dfs，思维）

题干：

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 nand 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：

给一个n*m的矩阵，'#'代表墙，'.'代表通路，给一个k，代表需要你将矩阵中k个通路改为X，但是要保证剩下的点依然联通，然后输出改变后的矩阵。

解题报告：

搜索。还有一种思路去搜索，那就是设有s个 ' . ' ，于是只搜s-k个点，然后就停止，最后把剩下的没搜的都变成 ' X ' 就好了。

AC代码1：（46ms）

#include<bits/stdc++.h>using namespace std;
int n,m,K;
char maze[550][550];
int vis[550][550];
int nx[4] = {0,1,0,-1};
int ny[4] = {1,0,-1,0};
void dfs(int x, int y) {int tx,ty;for(int k = 0; k<4; k++) {tx = x+nx[k];ty = y+ny[k];if(tx <1 || tx > n || ty < 1 || ty > m) continue;if(maze[tx][ty] == '#' || vis[tx][ty]!=0) continue;vis[tx][ty] = 1;dfs(tx,ty);}if(K>0) {K--;vis[x][y]=666;}
}
int main()
{cin>>n>>m>>K;for(int i = 1; i<=n; i++) {scanf("%s",maze[i]+1);}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.' && K > 0) dfs(i,j);}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(vis[i][j]!=666)printf("%c",maze[i][j]);elseprintf("X");}printf("\n");}return 0;}

WA代码2：（WA3了）

#include<bits/stdc++.h>using namespace std;
int n,m,K;
char maze[550][550];
int vis[550][550];
int nx[4] = {0,1,0,-1};
int ny[4] = {1,0,-1,0};
void dfs(int x, int y) {int tx,ty;for(int k = 0; k<4; k++) {tx = x+nx[k];ty = y+ny[k];if(tx <1 || tx > n || ty < 1 || ty > m) continue;if(maze[tx][ty] == '#' || vis[tx][ty]==1) continue;vis[tx][ty] = 1;dfs(tx,ty);}if(K>0) {K--;vis[x][y]=666;}
}
int main()
{cin>>n>>m>>K;for(int i = 1; i<=n; i++) {scanf("%s",maze[i]+1);}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.' && K > 0) dfs(i,j);}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(vis[i][j]!=666)printf("%c",maze[i][j]);elseprintf("X");}printf("\n");}return 0;}

AC代码3：（31ms）

#include<bits/stdc++.h>using namespace std;
int n,m,K;
char maze[550][550];
int vis[550][550];
int nx[4] = {0,1,0,-1};
int ny[4] = {1,0,-1,0};
void dfs(int x, int y) {int tx,ty;for(int k = 0; k<4; k++) {tx = x+nx[k];ty = y+ny[k];if(tx <1 || tx > n || ty < 1 || ty > m) continue;if(maze[tx][ty] == '#' || vis[tx][ty]==1) continue;vis[tx][ty] = 1;dfs(tx,ty);}if(K>0) {K--;maze[x][y] = 'X';}
}
int main()
{cin>>n>>m>>K;for(int i = 1; i<=n; i++) {scanf("%s",maze[i]+1);}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.' && K > 0) dfs(i,j);}}for(int i = 1; i<=n; i++) {printf("%s",maze[i]+1);printf("\n");}return 0;}

【Codeforces - 378C】Maze（dfs，思维）相关推荐

codeforces 378C MAZE
D - Maze Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u Submit St ...
Monopole Magnets CodeForces - 1345D（dfs+思维）
A monopole magnet is a magnet that only has one pole, either north or south. They don't actually exi ...
Last Theorem CodeForces - 1325F（dfs树找最大环+思维）
It's the year 5555. You have a graph, and you want to find a long cycle and a huge independent set, ...
Codeforces Round #606 (Div. 2, based on Technocup 2020 Elimination Round 4) dfs + 思维
传送门文章目录题意: 思路: 题意: 给一张图,求必须经过aaa点和bbb点的路径条数. 思路: 通过观察我们发现,这个路径无非就是x−>a−>b−>yx->a->b ...
CodeForces - 1293C NEKO's Maze Game(思维，水题)
题目链接:点击查看题目大意:给出一个2*n大小的矩阵,现在有m次操作,每次操作将某一个方格的状态置反,这里的每个方块都有两种状态,一种状态是可通行状态,另一种是不可通行状态,初始时所有方块都是可通行 ...
Codeforces Round #607 (Div. 2) E. Jeremy Bearimy dfs + 思维
传送门文章目录题意: 思路: 题意: 给你2∗k2*k2∗k个点的一棵树.定义GGG为任选kkk组不同的点,每组点的距离和的最小值.定义BBB为任选kkk组不同的点,每组点的距离和的最大值.让你求 ...
CodeForces - 1323B Count Subrectangles(思维)
题目链接:点击查看题目大意:给出一个数组 a 和数组 b 只由 0 和 1 构成,构造出矩阵 maze[ x ][ y ] = a[ x ] * b[ y ],显然maze矩阵同样只由 0 和 1 ...
Military Problem CodeForces 1006E （dfs序）
J - Military Problem CodeForces - 1006E 就是一道dfs序的问题给定一个树, 然后有q次询问. 每次给出u,k, 求以u为根的子树经过深搜的第k个儿子,如果一个 ...
codeforces 723D（DFS）
题目链接:http://codeforces.com/problemset/problem/723/D 题意:n*m的矩阵中,'*'代表陆地,'.'代表水,连在一起且不沿海的水形成湖泊.问最少填多少块 ...

【Codeforces - 378C】Maze（dfs，思维）

【Codeforces - 378C】Maze（dfs，思维）相关推荐

最新文章

热门文章