POJ 1979 Red and Black 基础性的DFS

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

45

59

6

13

题意：以‘@’为起始点，找出所有能到达的‘.’的个数，‘#’视为墙，不可走。（‘@’本身也算‘.’）

思路：见代码

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};char map[30][30];bool vis[30][30];int w,h,sum;int kx,ky;void dfs(int kx,int ky){    for(int i = 0; i < 4; i++)//搜索点的四个方向    {        int tx = kx + dir[i][0];        int ty = ky + dir[i][1];        if(!vis[tx][ty])//搜索到一个‘.’ ， 进行记录并以该点为起始点继续深搜        {            vis[tx][ty] = 1;//把搜索出来的‘.’记录，防止重复计算            sum++;            dfs(tx,ty);        }    }}int main( ){    while(cin>>w>>h&&w&&h)    {        getchar();        memset(vis,1,sizeof(vis));//把vis初始化为1，可无需设置边界，        for(int i = 1; i <= h; i++)        {            for(int j = 1; j <= w; j++)            {                cin>>map[i][j];                if('@'==map[i][j])                {                    kx = i;                    ky = j;                }                else if('.'==map[i][j])                {                    vis[i][j]=0;//输入的‘.’记录为0，意为可走                }            }            getchar();        }        sum = 1;        dfs(kx,ky);        cout<<sum<<endl;    }    return 0;}