ZOJ Problem Set - 2165 Red and Black-白红宇

ZOJ Problem Set - 2165 Red and Black

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发布时间：2019-06-09

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ZOJ Problem Set - 2165

Red and Black

Time Limit: 2 Seconds

Memory Limit: 65536 KB

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile

'#' - a red tile

'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9

....#.

.....#

......

#@...#

.#..#.

11 9

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

7 7

..#.#..

###.###

...@...

###.###

..#.#..

0 0

Sample Output

#include
     
      #include
      
       #include
       
        #define MAX 30using namespace std;char map[MAX][MAX];int count=0;int w,h;int dir[4][2]={
    {0,-1},{-1,0},{0,1},{1,0}};void dfs(int r,int c){//	cout<<"r: "<
        
         <<" c: "<
         
          <
          
           h||c>w) { return; } map[r][c]='#'; for(int i=0;i<4;i++) { if(map[r+dir[i][0]][c+dir[i][1]]=='.') { // map[r+dir[i][0]][c+dir[i][1]]='#'; count++; dfs(r+dir[i][0],c+dir[i][1]); // map[r+dir[i][0]][c+dir[i][1]]='.'; } }}int main(){ while(cin>>w>>h) { if(w==0||h==0)break; int sr,sc; count=0; memset(map,0,sizeof(map)); for(int i=1;i<=h;i++) { for(int j=1;j<=w;j++) { cin>>map[i][j]; if(map[i][j]=='@') { sr=i;sc=j; } } } map[sr][sc]='#'; //cout<<"Mark #1"<

转载于:https://www.cnblogs.com/jackwuyongxing/p/3366493.html

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