B - Red and Black
Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
题目大意：
就是说，对于一个h*w的网格，每次从‘@’开始，求最多能走多少个‘.’，每次只能够向四个方向进行扩展。
解题思路：
直接dfs，然后用book数组随时记录目前这个点有没有走过，如果这个点是‘.‘并且没有走过的话，那么我们就res++，由于是多组数据，每次输出结束后，记着memset,
复杂度：
O(4*w*h)
代码：
1 # include<cstdio>
2 # include<iostream>
3 # include<cstring>
4
5 using namespace std;
6
7 # define MAX 23
8
9 char grid[MAX][MAX];
10 int book[MAX][MAX];
11 int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1} };
12 int res;
13 int w,h;
14
15 int can_move ( int x,int y )
16 {
17 if ( x>=0&&x<h&&y>=0&&y<w )
18 {
19 if ( grid[x][y]==‘.‘&&book[x][y]==0 )
20 return 1;
21 }
22 return 0;
23 }
24
25 void dfs ( int x,int y )
26 {
27 book[x][y] = 1;
28 for ( int i = 0;i < 4;i++ )
29 {
30 int n_x = x+nxt[i][0], n_y = y+nxt[i][1];
31 if ( can_move ( n_x,n_y ) )
32 {
33 res++;
34 book[n_x][n_y] = 1;
35 dfs(n_x,n_y);
36 }
37 }
38 }
39
40 int main(void)
41 {
42 while ( scanf("%d%d",&w,&h)!=EOF )
43 {
44 res = 1;
45 if ( w==0&&h==0 )
46 break;
47 for ( int i = 0;i < h;i++ )
48 scanf("%s",grid[i]);
49 for ( int i = 0;i < h;i++ )
50 {
51 for ( int j = 0;j < w;j++ )
52 {
53 if ( grid[i][j]==‘@‘ )
54 {
55 book[i][j] = 1;
56 dfs(i,j);
57 }
58 }
59 }
60
61 printf("%d\n",res);
62 memset(book,0,sizeof(book));
63 }
64
65
66
67 return 0;
68 }