ZOJ2165 Red and Black(dfs)

Red and Black
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

4559613

翻译：
题目编号：ZOJ2165

详见：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2165

1.2题目描述：

有一个长方形的房间，房间里的地面上布满了正方形的瓷砖，瓷砖要么是红色，要么是黑色。一男子站在其中一块黑色的瓷砖上。男子可以向他四周的瓷砖上移动，但不能移动到红色的瓷砖上，只能在黑色的瓷砖上移动。

本题的目的就是要编写程序，计算他在这个房间里可以到达的黑色瓷砖的数量。

1.3输入描述

输入文件中包含多个测试数据。每个测试数据的第1 行为两个整数W和H，分别表示长方形房间里x方向和y方向上瓷砖的数目。W和H的值不超过20。接下来有H 行，每行有W个字符，每个字符代表了瓷砖的颜色，这些字符的取值及含义为：

1) ‘.’ － 黑色的瓷砖；

2) ‘#’ － 红色的瓷砖；

3) ‘@’ － 表示该位置为黑色瓷砖，且一名男子站在上面，注意每个测试数据中只有一个

‘@’符号。

输入文件中最后一行为两个0，代表输入文件结束。

1.4 输出描述

对输入文件中每个测试数据，输出占一行，为该男子从初始位置出发可以到达的黑色瓷砖的数目（包括他初始时所处的黑色瓷砖）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>#include<cstdlib>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const int maxn = 21;bool vis[maxn][maxn]; //标记是否走过int m, n;char oil[maxn][maxn];int dir[4][2] = {1,0, 0,1, 0,-1, -1,0}; //记录该点周围的方向void dfs(int x, int y){    vis[x][y] = 0;    int i;    int k, g;    for(i=0; i<4; i++){        k = x + dir[i][0];        g = y + dir[i][1];        //printf("%c    %d   %d      %d     \n",oil[k][g],vis[k][g],k,g);        if(oil[k][g]=='.' && vis[k][g]!=0 && k>=0 && k< n && g>=0 && g<m){             //   cout<<k<<"   "<<g<<endl;            dfs(k, g);        }    }}int main(){    int i, j, sum, e, t;    while(scanf("%d%d", &m, &n)!=EOF&& m+n)    {        sum = 0;        memset(vis, 1, sizeof(vis));        memset(oil, 0, sizeof(oil));        for(i=0; i<n; i++){            for(j=0; j<m; j++){                cin >> oil[i][j];                if(oil[i][j]=='@'){                    t=i;                    e=j;                }            }        }        dfs(t, e);        for(i=0; i<n; i++){            for(j=0; j<m; j++){                if(vis[i][j]==0){                    sum++;                }            }        }        cout << sum << endl;    }    return 0;}