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思维题|Maze（CodeForces - 377A ）（思维，广搜）

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A. Maze 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 【思维题|Maze（CodeForces - 377A ）（思维，广搜）】 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#.. ...# .#.#


题意理解：#代表墙.代表空想在点的地方插入K个X，剩下的点还是联通的。
解题思路：这主要就是一个逆向的思维,你只需要找到 . 有 n 个 ，那么要插入k 个X,也就是你要找到有n-k个 . 联通就行。再会广搜搜一下就可以了。
#include #include #include using namespace std; char map_[1010][1010]; struct s{ int x; int y; }a; struct qq{ int x; int y; }; int book[1010][1010]; int main() { int next[4][2]={{1,0},{-1,0},{0,-1},{0,1}}; int n,m,k; int t=0; scanf("%d%d%d",&n,&m,&k); for(int i=0; iq; while(!q.empty()) { q.pop(); } qq start; int flag=0; start.x=a.x; start.y=a.y; book[a.x][a.y]=1; q.push(start); while(!q.empty()) { if(cnt==t-k) { flag=1; break; } qq temp=q.front(); q.pop(); for(int p=0; p<4; p++) { qq tmp; tmp.x=temp.x+next[p][0]; tmp.y=temp.y+next[p][1]; if(tmp.x<0||tmp.y<0||tmp.x>n-1||tmp.y>m-1) { continue; } // printf("tmp.x=%d tmp; y=%d\n",tmp.x,tmp.y); if(map_[tmp.x][tmp.y]=='.'&&book[tmp.x][tmp.y]==0) { book[tmp.x][tmp.y]=1; q.push(tmp); cnt++; if(cnt==t-k) { flag=1; break; } } } if(flag==1) { break; } } for(int i=0; i





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