OpenJudge/Poj 1979 Red and Black / OpenJudge 2816 红与黑

1.链接地址：

http://bailian.openjudge.cn/practice/1979

http://poj.org/problem?id=1979

2.题目：

总时间限制:

1000ms

内存限制:

65536kB

描述

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.

输入

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.

输出

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).

样例输入

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

样例输出

45
59
6
13

来源

Japan 2004 Domestic

3.思路：

4.代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 
 6 int f(char **arr,const int i,const int j,int w,int h)
 7 {
 8     int res = 0;
 9     if(arr[i][j] == '.') 
10     {
11         res += 1; 
12         arr[i][j] = '#';
13         if(i > 0) res += f(arr,i - 1,j,w,h);
14         if(i < h - 1) res += f(arr,i + 1,j,w,h);
15         if(j > 0) res += f(arr,i,j - 1,w,h);
16         if(j < w - 1) res += f(arr,i,j + 1,w,h);
17     }
18     return res;
19 }
20 
21 
22 int main()
23 {
24     //freopen("C:\\Users\\wuzhihui\\Desktop\\input.txt","r",stdin);
25 
26     int i,j;
27 
28     int w,h;
29     //char ch;
30     while(cin>>w>>h)
31     {
32         if(w == 0 && h == 0) break;
33         //cin>>ch;
34 
35         char **arr = new char*[h];
36         for(i = 0; i < h; ++i) arr[i] = new char[w];
37 
38         for(i = 0; i < h; ++i)
39         {
40             for(j = 0; j < w; ++j)
41             {
42                 cin>>arr[i][j];
43             }
44             //cin>>ch;
45         }
46 
47         for(i = 0; i < h; ++i)
48         {
49             for(j = 0; j < w; ++j)
50             {
51                 if(arr[i][j] == '@')
52                 {
53                     arr[i][j] = '.';
54                     cout << f(arr,i,j,w,h) << endl;
55                     break;
56                 }
57             }
58             if(j < w) break;
59         }
60 
61         for(i = 0; i < h; ++i) delete [] arr[i];
62         delete [] arr;
63     }
64 
65     return 0;
66 }

 

转载于:https://www.cnblogs.com/mobileliker/p/3555011.html