HDU 1312 Red and Black【深搜】

 

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11531    Accepted Submission(s): 7177

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

 

 

Source

 

Asia 2004, Ehime (Japan), Japan Domestic

 

 

 

#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}};void dfs(int i,int j){int k,x,y;     cnt++;    map[i][j] = '#';    for(k = 0; k<4; k++)    {        x = i + to[k][0];        y = j + to[k][1];        if(x<n && y<m && x>=0 && y>=0 && map[x][y] == '.')//控制 x，y不让它越界             dfs(x,y);    }}int main(){    int i,j,fi,fj;    while(scanf("%d%d%",&m,&n))//横向为 n，纵向为 m     {    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] == '@')//找到刚开始人所在的位置                 {                    fi = i;                    fj = j;                }            }            getchar();        }        cnt = 0;        dfs(fi,fj);        printf("%d\n",cnt);    }    return 0;}