I - Red and Black DFS

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. 

InputThe 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) 
OutputFor 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
题目的大意就是搜图判断最多能遍历多少个".";简单dfs

#include<iostream>
#include<cstring>
using namespace std;
int w,h,ans;
char arr[25][25];
int mark[25][25];
int d[4][2]={{1,0},{0,1},{0,-1},{-1,0}};
void dfs(int x,int y)
{
    mark[x][y]=1;
    for(int i=0;i<4;i++){
        int dx=x+d[i][0];
        int dy=y+d[i][1];
        if(dx>=0&&dy>=0&&dx<h&&dy<w&&mark[dx][dy]==0&&arr[dx][dy]=='.'){
            ans++;
            mark[dx][dy]=1;
            dfs(dx,dy);
        }
    }
} 


int main()
{
    while(cin>>w>>h){
        memset(mark,0,sizeof(mark));
        ans=1;
        if(w==0&&h==0)
            break;
        for(int i=0;i<h;i++){
            scanf("%s",&arr[i]);
        }
        for(int i=0;i<h;i++)
            for(int j=0;j<w;j++){
                if(arr[i][j]=='@'){
//                    mark[i][j]=1;
                    dfs(i,j);
                }
            }    
            cout<<ans<<endl;
            
    }
    return 0;
}

BFS也可以写就是只要是x周围存在"."就加1,

#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
char arr[22][22];
int sa,ea;
int n,m;
int v[22][22]={0};
struct stu{
    int a,b;
//    int sum;
};

int a[4]={0,1,0,-1};
int b[4]={1,0,-1,0};

void bfs()
{
    int ans=0;
//    priority_queue<stu >que;
    queue<stu>que;
    stu q1;
    q1.a=sa;
    q1.b=ea;
//    q1.sum=1;
    v[sa][ea]=1;
    que.push(q1);
    
    while(que.size()){
        stu h;
//        h=que.top();
        h=que.front();
        que.pop();
        stu d;
        for(int i=0;i<4;i++){
            d.a=h.a+a[i];
            d.b=h.b+b[i];
            if(d.a>=0 && d.b>=0 && d.a<m && d.b<n&& v[d.a][d.b]!=1 && arr[d.a][d.b]!='#'){
                v[d.a][d.b]=1;
//                d.sum=h.sum+1;
//                cout<<d.sum<<endl;
                que.push(d);
//                ans=max(ans,d.sum);
                ans++;//只要加入到队列中就加一，因为加入到队列的一定是'.'
            }
        }
    }
    cout<<ans+1<<endl;
}

int main(){

    while(cin>>n>>m)
    {
        memset(v,0,sizeof(v));
        
        if(n==0&&m==0)
            break;
            
        for(int i=0;i<m;i++){
            scanf("%s",&arr[i]);
        }
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(arr[i][j]=='@'){
                    sa=i;
                    ea=j;
                }
            }
        }
        
        bfs();
    }
    return 0;
}