[DFS] HOJ 1797 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).

Sample Input
6 9
….#.
…..#
……
……
……
……
……
#@…#
.#..#.
11 9
.#………
.#.#######.
.#.#…..#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#…….#.
.#########.
………..
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
/###.###
…@…
/###.###
..#.#..
..#.#..
0 0a
Sample Output
45
59
6
13

题意:
一个屋子里有黑红两色瓷砖,一个男人站在黑砖上,他可以向上下左右的黑色瓷砖走,计算最多可以走多少瓷砖。

//已AC代码#include<bits/stdc++.h>using namespace std;int w,h;char a[100][100];int vis[100][100];int beginx,beginy;int ans;int check(int x,int y){    if(x>=0&&x<=h&&y>=0&&y<=w&&a[x][y]=='.'&&!vis[x][y])    return 1;    else    return 0;}int dfs(int beginx,int beginy){    ++ans;    int x=beginx;    int y=beginy;    vis[x][y]=1;    if(check(x-1,y)&&dfs(x-1,y))    {    }    if(check(x+1,y)&&dfs(x+1,y))    {    }    if(check(x,y-1)&&dfs(x,y-1))    {    }    if(check(x,y+1)&&dfs(x,y+1))    {    }}int main(){    while(~scanf("%d%d",&w,&h))    {        ans=0;        beginx=beginy=0;        memset(a,'#',sizeof(a));        memset(vis,0,sizeof(vis));        if(w==0&&h==0)            break;        for(int i=0;i<h;i++)        {            for(int j=0;j<w;j++)            {                cin>>a[i][j];                if(a[i][j]=='@')                {                    beginx=i;                    beginy=j;                    vis[i][j]=0;                }            }        }        dfs(beginx,beginy);        cout<<ans<<endl;    }}