Red and Black
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice OpenJ_Bailian 1979
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.
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)
The end of the input is indicated by a line consisting of two zeros.
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
题意：
给出一个字符矩阵表示一个房间内的情况，‘.’表示黑砖、‘#’表示红砖、‘@’表示一个人最初站立的位置。这个人不能走上红砖，问这个人能到达的地砖总数是多少。
输入：
多组数据，首先输入矩阵的列数和行数，然后输入矩阵。以列数与行数都为0表示输入的结束。
输出：
能到达的地砖总数。
分析：
使用深度优先搜索。
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 int dx[4] = {1,0,-1,0},dy[4] = {0,1,0,-1}; 8 int W,H; 9 char G[24][24]; 10 int cnt; 11 void dfs(int x,int y){ 12 cnt++; G[x][y] = '#'; 13 for(int i = 0 ;i < 4 ; i++){ 14 int nx = x + dx[i],ny = y + dy[i]; 15 if(nx >= 0 && nx < H && ny >= 0 && ny < W && G[nx][ny] != '#') 16 dfs(nx,ny); 17 } 18 } 19 int main(){ 20 while(scanf("%d%d",&W,&H) == 2 && W){ 21 cnt = 0; 22 int sx,sy; 23 getchar(); 24 for(int i = 0 ; i < H ; i++){ 25 for(int j = 0 ; j < W ; j++){ 26 scanf("%c",&G[i][j]); 27 if(G[i][j] == '@'){ 28 sx = i; 29 sy = j; 30 } 31 } 32 getchar(); 33 } 34 //for(int i = 0 ;i < H ;i++){ for(int j = 0 ;j < W; j++) printf("%c",G[i][j]);printf(" ");} 35 dfs(sx,sy); printf("%d ",cnt); 36 } 37 return 0; 38 }