hdoj -1312 Red and Black

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). 

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1312

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace::std;

int s_i,s_j,num;
char maps[25][25];
int maps_[25][25]={0},a[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
int W,H;
void dfs(int n,int m)
{
    for(int i=0;i<4;i++)
    {
        int x = n + a[i][0];
        int y = m + a[i][1];
        if(x>=0 && y>=0 && x<H && y<W && maps_[x][y] != 1 && maps[x][y] != '#')
        {
            num++;
            maps_[x][y] = 1;
            dfs(x,y);
        }
    }
}
int main()
{

    int i,j;
    while(scanf("%d %d",&W,&H) && W != 0 && H != 0)
    {
    memset(maps,0,sizeof(maps));   //每次输入清空地图和标记
    memset(maps_,0,sizeof(maps_));
    for(i=0;i<H;i++)
    {
        scanf("%s",&maps[i]);
        for(j =0;j<W;j++)
        {
            if(maps[i][j] == '@')  //找到起始点
            {
                s_i = i;
                s_j = j;
                break;
            }
        }
    }
    num = 1;
    maps_[s_i][s_j] = 1;  //标记起始点
    dfs(s_i,s_j);
    printf("%d\n",num);
    }
    return 0;
}