Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7328 Accepted Submission(s): 4585
Problem Description
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.
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)
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
45
59
6
13
Source
Recommend
1 //0MS 256K 748 B C++ 2 /* 3 4 题意: 5 求与'@'连在一起的'.' 的数量 6 7 DFS: 8 基础dfs.. 9 10 */ 11 #include<stdio.h> 12 #include<string.h> 13 char g[25][25]; 14 int mov[4][2]={0,1,1,0,0,-1,-1,0}; 15 int n,m,s,e; 16 int cnt; 17 void dfs(int x,int y) 18 { 19 for(int i=0;i<4;i++){ 20 int tx=x+mov[i][0]; 21 int ty=y+mov[i][1]; 22 if(tx>=0 && tx<n && ty>=0 && ty<m && g[tx][ty]=='.'){ 23 cnt++; 24 g[tx][ty]='#'; 25 dfs(tx,ty); 26 } 27 } 28 } 29 int main(void) 30 { 31 while(scanf("%d%d",&m,&n),n+m) 32 { 33 for(int i=0;i<n;i++){ 34 scanf("%s",g[i]); 35 for(int j=0;j<m;j++) 36 if(g[i][j]=='@'){ 37 s=i,e=j; 38 } 39 } 40 cnt=1; 41 g[s][e]='#'; 42 dfs(s,e); 43 printf("%d ",cnt); 44 } 45 return 0; 46 }