POJ 1979 Red and Black（DFS）

Red and Black

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

题目大意：一个房间里面铺满了红色（用#表示）和黑色（用.表示）的地板，有一个人在房间里走动，他只能走在黑色地板上，不能走在红色地板上，这个人一开始站在某一块黑色地板上，求他能走过的黑色地板的数目。

解题思路：DFS求连通块，要算上最开始的那一块地板，注意输入的行和列是反着的。

代码如下：

#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxw = 25;const int maxh = 25;char room[maxh][maxw];int dir[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};int w,h;int ans;void dfs(int x,int y){    room[x][y] = '#';    ans++;    for(int i = 0;i < 4;i++){        int nx = x + dir[i][0];        int ny = y + dir[i][1];        if(0 <= nx && nx < h && 0 <= ny && ny < w && room[nx][ny] != '#'){            dfs(nx,ny);        }    }    return ;}int main(){    while(scanf("%d %d",&w,&h) != EOF && (w || h)){        ans = 0;        for(int i = 0;i < h;i++){            scanf("%s",room[i]);        }        for(int i = 0;i < h;i++){            for(int j = 0;j < w;j++){                if(room[i][j] == '@'){                    dfs(i,j);                }            }        }        printf("%d\n",ans);    }    return 0;}