poj1979 DFS

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 34961 Accepted: 18920

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

搜索的水题，注意字符输入时对回车键的处理。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;char g[25][25];int result;int n,m;void dfs(int x,int y){    if(x<=m-1&&x>=0&&y<=n-1&&y>=0&&g[x][y]=='.')    {        result++;        g[x][y]='#';    }    else        return;    dfs(x-1,y);    dfs(x,y-1);    dfs(x,y+1);    dfs(x+1,y);}int main(){    int p,q;    while(scanf("%d%d",&n,&m))    {        if(n==0&&m==0)        break;        result=0;        memset(g,0,sizeof(g));        for(int i = 0 ; i < m ; i ++)        {            getchar();            for(int j = 0; j < n;j ++)            {                scanf("%c",&g[i][j]);                if(g[i][j]=='@')                {                    p=i;q=j;                    g[i][j]='.';                }            }        }        dfs(p,q);        printf("%d\n",result);    }    return 0;}