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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(gè) X 且保持所有的 '.' 保持貫通
====================================================================================================================================================
怎么樣才能使插入的 X 沒(méi)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(gè)點(diǎn)一路走到底，直到不能走為止，那么這個(gè)末端放上X一定不會(huì)阻斷其它'.'的連貫
如果所有末端 插入完畢之后，還有X沒(méi)有放入，那么末端之后緊挨著的點(diǎn)就變成了末端，根據(jù)DFS走到末端后會(huì)原路返回，那么剩下的X跟著原路返回時(shí)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(gè)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(dāng)遍歷到底了之后，即循環(huán)了四次無(wú)法繼續(xù)走下去時(shí) 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(zhuǎn)載于:https://www.cnblogs.com/darkboy/p/9415934.html

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