Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13090 Accepted Submission(s): 8116
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
//比较基础的搜索题， 但是做了很久。→ → ！！
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #define M 30 5 using namespace std; 6 7 int ans; 8 int x, y; 9 char map[M][M]; 10 int dx[4]={0, 0, 1, -1}; 11 int dy[4]={1, -1, 0, 0}; 12 int a, b; 13 void dfs(int x, int y) 14 { 15 int m, n; 16 ans++; 17 map[x][y] = '@'; 18 //printf("%d %d ", x, y); 19 for(int i = 0; i < 4; i++) // 20 { 21 m = x + dx[i]; 22 n = y + dy[i]; 23 //printf("%d %d ", m, n); 24 if(m >= 0 && m < b && n >= 0 && n < a && map[m][n] == '.') //限制边界。 25 { 26 // printf("1"); 27 // printf(" %d %d ", m, n); 28 dfs(m, n); 29 } 30 } 31 } 32 33 int main() 34 { 35 int i, j; 36 while(~scanf("%d %d", &a, &b), a|b) 37 { 38 ans = 0; 39 for(i=0; i<b; i++) 40 { 41 for(j=0; j<a; j++) 42 { 43 cin >> map[i][j]; 44 if(map[i][j] == '@') 45 { 46 x = i; 47 y = j; 48 } 49 } 50 } 51 //printf("%d %d ", x, y); 52 dfs(x, y); 53 printf("%d ", ans); 54 } 55 return 0; 56 }
注意：
1 void dfs(int x, int y) 2 { 3 int nx, ny; 4 ant++; 5 map[x][y] = '@'; 6 for(int i = 0; i < 4; i++) 7 { 8 nx = x + acx[i]; 9 ny = y + acy[i]; 10 if(nx >= 0 && nx < m && ny >= 0 && ny < n && map[nx][ny] == '.') 11 { 12 dfs(nx, ny); 13 } 14 } 15 }