ZOJ 2165 Red and Black

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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)

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

题目大意：

给出两个数m和n，代表n行m列，都不超过20，然后是n行m列的图，包括‘.‘ , ‘#‘ , ‘@‘3个字符。

@代表初始位置，‘.‘代表通路，‘#’代表墙，求最大可到达的 ‘.‘ 的数量 ‘@‘算一个

BFS：

即求最大连通块问题 不一定非要用队列 一维数组一样可以解决

 1 #include<stdio.h>
 2 #include<string.h>
 3 struct node
 4 {
 5    int x,y;
 6 }q[410];
 7 
 8 struct node P, N;
 9 int flag[25][25];
10 int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
11 char str[25][25];
12 
13 int main()
14 {
15      int c, r, i, j, front, rear;
16      while(scanf("%d%d",&c,&r)!=EOF, c + r){
17         memset(flag, 0, sizeof(flag));
18         for(i = 0; i < r; i++)
19             scanf("%s", str[i]);
20 
21         for(i = 0; i < r; i++){
22             for(j=0;j<c;j++)
23                 if(str[i][j] == ‘@‘) break;
24             if(str[i][j] == ‘@‘) break;
25         }
26         N.x = i;
27         N.y = j;
28         flag[i][j] = 1;
29         q[0] = N;
30         front = 0;
31         rear = 1;
32 
33         while(front < rear){
34             N = q[front++];
35             for(i = 0; i < 4; i++){
36                 int tx = N.x + dir[i][0];
37                 int ty = N.y + dir[i][1];
38                 if(tx >= 0 && tx < r && ty >= 0&& ty < c && flag[tx][ty] == 0 && str[tx][ty] == ‘.‘){
39                     P.x = tx;
40                     P.y = ty;
41                     q[rear++] = P;
42                     flag[tx][ty] = 1;
43                 }
44             }
45         }
46         printf("%d\n",rear);
47      }
48      return 0;
49 }

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ZOJ 2165 Red and Black

标签：des   style   blog   color   os   strong   io   for   

原文地址：http://www.cnblogs.com/zzy9669/p/3884439.html