OpenJ_Bailian 1979

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

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 }

View Code