POJ1979 - Red and Black

题目：http://poj.org/problem?id=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

4559613

分析：

BFS。

代码：

#include <iostream>#include <queue>#include <cstring>int dx[] = {-1, 1, 0, 0};int dy[] = {0, 0, -1, 1};int main(){char map[22][22];bool visited[22][22];int W, H;std::pair<int, int> start;while (true){std::cin >> W >> H;if(W == 0 && H == 0)break;memset(map, '#', sizeof(map));memset(visited, false, sizeof(visited));for(int i = 0; i < H; ++i){for(int j = 0; j < W; ++j){std::cin >> map[i][j];if(map[i][j] == '@'){start = std::make_pair(i, j);}}}std::queue<std::pair<int, int> > q;q.push(start);visited[start.first][start.second] = true;int count = 1;while (!q.empty()){int x = q.front().first;int y = q.front().second;q.pop();  for(int i = 0; i < 4; ++i){int newx = x + dx[i];int newy = y + dy[i];if(newx >= 0 && newx < H && newy >= 0 && newy < W){if(!visited[newx][newy] && map[newx][newy] == '.'){q.push(std::make_pair(newx, newy));visited[newx][newy] = true;++count;}}}}std::cout << count << std::endl;}return 0;}

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