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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?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#..
...#
.#.#

題意：
插入 k 個 X 且保持所有的 '.' 保持貫通
====================================================================================================================================================
怎么樣才能使插入的 X 沒有阻斷路的連通，，想一想DFS，不撞南墻不回頭的理論，只要沿著一個點一路走到底，直到不能走為止，那么這個末端放上X一定不會阻斷其它'.'的連貫
如果所有末端 插入完畢之后，還有X沒有放入，那么末端之后緊挨著的點就變成了末端，根據DFS走到末端后會原路返回，那么剩下的X跟著原路返回時插入就行了。 ====================================================================================================================================================
代碼：
1 #include<bits/stdc++.h> 2 using namespace std; 3 char Map[510][510]; 4 bool book[510][510]; 5 int n,m,k; 6 void dfs(int x,int y) 7 { 8 if(x<0||x>=n||y<0||y>=m) return; 9 if(Map[x][y]!='.'||book[x][y]==1) return ; 10 book[x][y]=1; 11 for(int i=1;i<=4;i++) 12 { //四個方向 13 if(i==1) dfs(x+1,y); 14 if(i==2) dfs(x-1,y); 15 if(i==3) dfs(x,y+1); 16 if(i==4) dfs(x,y-1); 17 } 18 //當遍歷到底了之后，即循環了四次無法繼續走下去時 19 if(k!=0) 20 Map[x][y]='X',k--; 21 } 22 int main() 23 { 24 scanf("%d%d%d",&n,&m,&k); 25 for(int i=0;i<n;++i) 26 scanf("%s",Map[i]); 27 28 for(int i=0;i<n;++i) 29 { 30 for(int j=0;j<m;j++) 31 { 32 dfs(i,j); 33 if(k==0) break; 34 } 35 if(k==0) break; 36 } 37 for(int i=0;i<n;i++) 38 puts(Map[i]); 39 return 0; 40 }

轉載于:https://www.cnblogs.com/darkboy/p/9415934.html

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