題干：

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#.. ...# .#.#

題目大意：

給一個n*m的矩陣，'#'代表墻，'.'代表通路，給一個k，代表需要你將矩陣中k個通路改為X，但是要保證剩下的點依然聯通，然后輸出改變后的矩陣。

解題報告：

? ? ?搜索。還有一種思路去搜索，那就是 設有s個 ' . ' ，于是只搜s-k個點，然后就停止，最后把剩下的沒搜的都變成 '?X ' 就好了。

AC代碼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代碼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代碼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;}

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