DFS 深搜 HDU 1312

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Problem 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)

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
这样的题目很简单
#include <stdio.h>#include <string.h>int n,m,cnt;char map[30][30];int to[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};         //只有四个方向void dfs(int i,int j){    cnt++;    map[i][j] = '#';                                //走过的直接变成#    for(int k = 0; k<4; k++)    {        int x = i+to[k][0];        int y = j+to[k][1];        if(x<n && y<m && x>=0 && y>=0 && map[x][y] == '.')            dfs(x,y);    }    return;}int main(){    int i,j,fi,fj;    while(~scanf("%d%d%*c",&m,&n))    {        if(m == 0 && n == 0)            break;        for(i = 0; i<n; i++)        {            for(j = 0; j<m; j++)            {                scanf("%c",&map[i][j]);                if(map[i][j] == '@')                {                    fi = i;                    fj = j;                }            }            getchar();        }        cnt = 0;        //map[i][j] = '#';        dfs(fi,fj);        printf("%d\n",cnt);    }    return 0;}

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