POJ 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19254		Accepted: 10252

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

 

一道基础的深搜题，一个人从给定的位置开始走，可以朝上下左右四个方向，只能走黑色的格子，问这个人一共可以走多少个格子

只要从起始位置开始深搜就可以了

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int w,h,start_x,start_y,ans;
bool dfsed[30][30];
char g[30][30];

void dfs(int x,int y)
{
    if(g[x-1][y]=='.'&&(!dfsed[x-1][y]))
    {
        dfsed[x-1][y]=true;
        dfs(x-1,y);
        ans++;
    }
    if(g[x+1][y]=='.'&&(!dfsed[x+1][y]))
    {
        dfsed[x+1][y]=true;
        dfs(x+1,y);
        ans++;
    }
    if(g[x][y-1]=='.'&&(!dfsed[x][y-1]))
    {
        dfsed[x][y-1]=true;
        dfs(x,y-1);
        ans++;
    }
    if(g[x][y+1]=='.'&&(!dfsed[x][y+1]))
    {
        dfsed[x][y+1]=true;
        dfs(x,y+1);
        ans++;
    }
}

int main()
{
    while(scanf("%d %d",&w,&h)==2)
    {
        getchar();
        if(w==0&&h==0)
            break;
        for(int i=1;i<=h;i++)
            gets(&g[i][1]);
        for(int i=0;i<=h+1;i++)
            g[i][0]=g[i][w+1]='#';
        for(int i=1;i<=w+1;i++)
            g[0][i]=g[h+1][i]='#';
        for(int i=1;i<=h;i++)
            for(int j=1;j<=w;j++)
                if(g[i][j]=='@')
                {
                    start_x=i;
                    start_y=j;
                }
        ans=1;
        memset(dfsed,false,sizeof(dfsed));
        dfs(start_x,start_y);
        cout<<ans<<endl;
    }

    return 0;
}